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Integral $intfrac{1}{1+x^3}dx$
$begingroup$
Calculate$$intfrac{1}{1+x^3}dx$$
After calculating the partial fractions I got:
$$frac{1}{3}intfrac{1}{x+1}dx+frac{1}{3}intfrac{2-x}{x^2-x+1}dx=frac{1}{3}ln(x+1)+frac{1}{3}intfrac{2-x}{x^2-x+1}dx$$
I have no idea on how to proceed. Am I missing a substitution or something?
integration indefinite-integrals
edited Jan 29 at 23:49
Zacky
7,81511062
asked Jan 29 at 23:34
VakiPitsiVakiPitsi
1998
$endgroup$
add a comment |
$begingroup$
Calculate$$intfrac{1}{1+x^3}dx$$
After calculating the partial fractions I got:
$$frac{1}{3}intfrac{1}{x+1}dx+frac{1}{3}intfrac{2-x}{x^2-x+1}dx=frac{1}{3}ln(x+1)+frac{1}{3}intfrac{2-x}{x^2-x+1}dx$$
I have no idea on how to proceed. Am I missing a substitution or something?
integration indefinite-integrals
edited Jan 29 at 23:49
Zacky
7,81511062
asked Jan 29 at 23:34
VakiPitsiVakiPitsi
1998
$endgroup$
add a comment |
$begingroup$
Calculate$$intfrac{1}{1+x^3}dx$$
After calculating the partial fractions I got:
$$frac{1}{3}intfrac{1}{x+1}dx+frac{1}{3}intfrac{2-x}{x^2-x+1}dx=frac{1}{3}ln(x+1)+frac{1}{3}intfrac{2-x}{x^2-x+1}dx$$
I have no idea on how to proceed. Am I missing a substitution or something?
integration indefinite-integrals
edited Jan 29 at 23:49
Zacky
7,81511062
asked Jan 29 at 23:34
VakiPitsiVakiPitsi
1998
$endgroup$
Calculate$$intfrac{1}{1+x^3}dx$$
After calculating the partial fractions I got:
$$frac{1}{3}intfrac{1}{x+1}dx+frac{1}{3}intfrac{2-x}{x^2-x+1}dx=frac{1}{3}ln(x+1)+frac{1}{3}intfrac{2-x}{x^2-x+1}dx$$
I have no idea on how to proceed. Am I missing a substitution or something?
integration indefinite-integrals
integration indefinite-integrals
edited Jan 29 at 23:49
Zacky
7,81511062
asked Jan 29 at 23:34
VakiPitsiVakiPitsi
1998
edited Jan 29 at 23:49
Zacky
7,81511062
asked Jan 29 at 23:34
VakiPitsiVakiPitsi
1998
edited Jan 29 at 23:49
Zacky
7,81511062
edited Jan 29 at 23:49
Zacky
7,81511062
edited Jan 29 at 23:49
Zacky
7,81511062
7,81511062
asked Jan 29 at 23:34
VakiPitsiVakiPitsi
1998
asked Jan 29 at 23:34
VakiPitsiVakiPitsi
1998
asked Jan 29 at 23:34
VakiPitsiVakiPitsi
1998
1998
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
To integrate $frac{2- x}{x^2- x+ 1}$, complete the square in the denominator: $int frac{2- x}{x^2- x+ 1}dx= intfrac{2- x}{x^2- x+ frac{1}{4}-frac{1}{4}+ 1}dx= intfrac{2-x}{left(x- frac{1}{2}right)+ frac{3}{4}}dx$.
Now let $u= x- frac{1}{2}$ so that $du= dx$ and $x= u+ frac{1}{2}$ and $2- x= frac{3}{2}- u$ and $dx= -dy.
$int frac{2- x}{x^2- x+ 1}dx= $$-int frac{frac{3}{2}- u}{u^2+ frac{3}{4}}du= $$-frac{3}{2}intfrac{du}{u^2+ frac{3}{4}}+ intfrac{u du}{u^2+ frac{3}{4}}$.
answered Jan 30 at 0:04
user247327user247327
11.6k1516
$endgroup$
$begingroup$
I still have some difficulty to calculate $int frac{1}{u^2+frac{3}{4}}du$. But otherwise very explanatory answer thank you
$endgroup$
– VakiPitsi
Jan 30 at 0:27
1
$begingroup$
Multiply both numerator and denominator by $frac{4}{2}$ to get $frac{4}{3}intfrac{1}{frac{4}{3}u^2+ 1}du$. Now let $v= frac{2}{sqrt{3}}u$ so that $dv= frac{2}{sqrt{3}}du$, $du= frac{sqrt{3}}{2}dv$, and the integral becomes $frac{2}{sqrt{3}}intfrac{1}{v^2+ 1}dv$. That integral is an arctangent.
$endgroup$
– user247327
Jan 30 at 12:57
add a comment |
$begingroup$
Hint: You can break up your second calculated indefinite integral into two components as so:
$$frac{2-x}{x^2-x+1} = -frac{1}{2}(frac{2x -1}{x^2-x+1}) + frac{3/2}{(x-frac{1}{2})^2 + 3/4}$$
Note that $x^2-x+1 = (x - frac{1}{2})^2 + frac{3}{4}$
answered Jan 29 at 23:44
HyperionHyperion
702111
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add a comment |
$begingroup$
Alternative approach: Partial fractions.
Recall that for any complex $z$, and any $n=1,2,...$
$$z^{1/n}=|z|^{1/n}expleft[frac{i}{n}(2pi k+arg z)right],qquad k=0,1,...,n-1$$
Then plug in $z=-1$ and $n=3$ to see that $arg z=arg(-1)=pi$ so that in fact,
$$1+x^3=prod_{k=0}^2left(x-expfrac{ipi(2k+1)}3right)$$
So, letting $r_k=expfrac{ipi(2k+1)}3$,
$$frac1{1+x^3}=prod_{k=0}^2frac1{x-r_k}$$
Look! That's a thing we can do partial fractions on! To do so, we say that
$$prod_{k=0}^2frac1{x-r_k}=sum_{k=0}^2frac{b(k)}{x-r_k}$$
$$left(prod_{a=0}^2(x-r_a)right)prod_{k=0}^2frac1{x-r_k}=left(prod_{a=0}^2(x-r_a)right)sum_{k=0}^2frac{b(k)}{x-r_k}$$
$$1=sum_{k=0}^2frac{b(k)}{x-r_k}prod_{a=0}^2(x-r_a)$$
$$1=sum_{k=0}^2b(k)prod_{a=0\ aneq k}^2(x-r_a)$$
then for any $j=0,1,2$, we may plug in $x=r_j$ and notice that all the terms of the sum vanish except for the case $k=j$, which gives
$$1=b(j)prod_{a=0\ aneq j}^2(r_j-r_a)$$
$$b(j)=prod_{a=0\ aneq j}^2frac1{r_j-r_a}$$
Which is an explicit formula for our partial fractions coefficients. Anyway, we may now integrate:
$$I=intfrac{mathrm dx}{1+x^3}=intsum_{k=0}^2frac{b(k)}{x-r_k}mathrm dx$$
$$I=sum_{k=0}^2b(k)intfrac{mathrm dx}{x-r_k}$$
Which is very easily shown to be
$$I=sum_{k=0}^{2}b(k)lnleft|x-r_kright|$$
$$I=b(0)lnleft|x-r_0right|+b(1)lnleft|x-r_1right|+b(2)lnleft|x-r_2right|$$
And since $b(k)=prodlimits_{a=0\ aneq k}^2frac1{r_k-r_a}$, we have that
$$
b(0)=frac1{(r_0-r_1)(r_0-r_2)}\
b(1)=frac1{(r_1-r_0)(r_1-r_2)}\
b(2)=frac1{(r_2-r_0)(r_2-r_1)}$$
And from $exp(itheta)=costheta+isintheta$,
$$
r_0=expfrac{ipi}3=frac{1+isqrt3}2\
r_1=expfrac{3ipi}3=-1\
r_2=expfrac{5ipi}3=frac{1-isqrt3}2
$$
So
$$
b(0)=-frac16(1+isqrt3)\
b(1)=frac16(1+isqrt3)\
b(2)=frac16(-1+isqrt3)
$$
And finally
$$I=-frac{1+isqrt3}6lnleft|x-frac{1+isqrt3}2right|+frac{1+isqrt3}6lnleft|x+1right|+frac{-1+isqrt3}6lnleft|x+frac{-1+isqrt3}2right|+C$$
In fact, using the same method, it can be shown that
$$intfrac{mathrm dx}{1+x^n}=sum_{k=0}^{n-1}lnleft|x-expfrac{ipi(2k+1)}{n}right|prod_{a=0\aneq k}^{n-1}frac1{expfrac{ipi(2k+1)}{n}-expfrac{ipi(2a+1)}{n}}$$
answered Jan 30 at 1:47
clathratusclathratus
5,0641439
$endgroup$
1
$begingroup$
Before I got halfway through reading the solution, I knew it was yours!
$endgroup$
– user150203
Jan 30 at 6:42
$begingroup$
@DavidG Haha thanks. I can usually tell which solutions are yours too :)
$endgroup$
– clathratus
Jan 30 at 21:36
add a comment |
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3 Answers
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3 Answers
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$begingroup$
To integrate $frac{2- x}{x^2- x+ 1}$, complete the square in the denominator: $int frac{2- x}{x^2- x+ 1}dx= intfrac{2- x}{x^2- x+ frac{1}{4}-frac{1}{4}+ 1}dx= intfrac{2-x}{left(x- frac{1}{2}right)+ frac{3}{4}}dx$.
Now let $u= x- frac{1}{2}$ so that $du= dx$ and $x= u+ frac{1}{2}$ and $2- x= frac{3}{2}- u$ and $dx= -dy.
$int frac{2- x}{x^2- x+ 1}dx= $$-int frac{frac{3}{2}- u}{u^2+ frac{3}{4}}du= $$-frac{3}{2}intfrac{du}{u^2+ frac{3}{4}}+ intfrac{u du}{u^2+ frac{3}{4}}$.
answered Jan 30 at 0:04
user247327user247327
11.6k1516
$endgroup$
$begingroup$
I still have some difficulty to calculate $int frac{1}{u^2+frac{3}{4}}du$. But otherwise very explanatory answer thank you
$endgroup$
– VakiPitsi
Jan 30 at 0:27
1
$begingroup$
Multiply both numerator and denominator by $frac{4}{2}$ to get $frac{4}{3}intfrac{1}{frac{4}{3}u^2+ 1}du$. Now let $v= frac{2}{sqrt{3}}u$ so that $dv= frac{2}{sqrt{3}}du$, $du= frac{sqrt{3}}{2}dv$, and the integral becomes $frac{2}{sqrt{3}}intfrac{1}{v^2+ 1}dv$. That integral is an arctangent.
$endgroup$
– user247327
Jan 30 at 12:57
add a comment |
$begingroup$
To integrate $frac{2- x}{x^2- x+ 1}$, complete the square in the denominator: $int frac{2- x}{x^2- x+ 1}dx= intfrac{2- x}{x^2- x+ frac{1}{4}-frac{1}{4}+ 1}dx= intfrac{2-x}{left(x- frac{1}{2}right)+ frac{3}{4}}dx$.
Now let $u= x- frac{1}{2}$ so that $du= dx$ and $x= u+ frac{1}{2}$ and $2- x= frac{3}{2}- u$ and $dx= -dy.
$int frac{2- x}{x^2- x+ 1}dx= $$-int frac{frac{3}{2}- u}{u^2+ frac{3}{4}}du= $$-frac{3}{2}intfrac{du}{u^2+ frac{3}{4}}+ intfrac{u du}{u^2+ frac{3}{4}}$.
answered Jan 30 at 0:04
user247327user247327
11.6k1516
$endgroup$
$begingroup$
I still have some difficulty to calculate $int frac{1}{u^2+frac{3}{4}}du$. But otherwise very explanatory answer thank you
$endgroup$
– VakiPitsi
Jan 30 at 0:27
1
$begingroup$
Multiply both numerator and denominator by $frac{4}{2}$ to get $frac{4}{3}intfrac{1}{frac{4}{3}u^2+ 1}du$. Now let $v= frac{2}{sqrt{3}}u$ so that $dv= frac{2}{sqrt{3}}du$, $du= frac{sqrt{3}}{2}dv$, and the integral becomes $frac{2}{sqrt{3}}intfrac{1}{v^2+ 1}dv$. That integral is an arctangent.
$endgroup$
– user247327
Jan 30 at 12:57
add a comment |
$begingroup$
To integrate $frac{2- x}{x^2- x+ 1}$, complete the square in the denominator: $int frac{2- x}{x^2- x+ 1}dx= intfrac{2- x}{x^2- x+ frac{1}{4}-frac{1}{4}+ 1}dx= intfrac{2-x}{left(x- frac{1}{2}right)+ frac{3}{4}}dx$.
Now let $u= x- frac{1}{2}$ so that $du= dx$ and $x= u+ frac{1}{2}$ and $2- x= frac{3}{2}- u$ and $dx= -dy.
$int frac{2- x}{x^2- x+ 1}dx= $$-int frac{frac{3}{2}- u}{u^2+ frac{3}{4}}du= $$-frac{3}{2}intfrac{du}{u^2+ frac{3}{4}}+ intfrac{u du}{u^2+ frac{3}{4}}$.
answered Jan 30 at 0:04
user247327user247327
11.6k1516
$endgroup$
To integrate $frac{2- x}{x^2- x+ 1}$, complete the square in the denominator: $int frac{2- x}{x^2- x+ 1}dx= intfrac{2- x}{x^2- x+ frac{1}{4}-frac{1}{4}+ 1}dx= intfrac{2-x}{left(x- frac{1}{2}right)+ frac{3}{4}}dx$.
Now let $u= x- frac{1}{2}$ so that $du= dx$ and $x= u+ frac{1}{2}$ and $2- x= frac{3}{2}- u$ and $dx= -dy.
$int frac{2- x}{x^2- x+ 1}dx= $$-int frac{frac{3}{2}- u}{u^2+ frac{3}{4}}du= $$-frac{3}{2}intfrac{du}{u^2+ frac{3}{4}}+ intfrac{u du}{u^2+ frac{3}{4}}$.
answered Jan 30 at 0:04
user247327user247327
11.6k1516
answered Jan 30 at 0:04
user247327user247327
11.6k1516
answered Jan 30 at 0:04
user247327user247327
11.6k1516
answered Jan 30 at 0:04
user247327user247327
11.6k1516
11.6k1516
$begingroup$
I still have some difficulty to calculate $int frac{1}{u^2+frac{3}{4}}du$. But otherwise very explanatory answer thank you
$endgroup$
– VakiPitsi
Jan 30 at 0:27
1
$begingroup$
Multiply both numerator and denominator by $frac{4}{2}$ to get $frac{4}{3}intfrac{1}{frac{4}{3}u^2+ 1}du$. Now let $v= frac{2}{sqrt{3}}u$ so that $dv= frac{2}{sqrt{3}}du$, $du= frac{sqrt{3}}{2}dv$, and the integral becomes $frac{2}{sqrt{3}}intfrac{1}{v^2+ 1}dv$. That integral is an arctangent.
$endgroup$
– user247327
Jan 30 at 12:57
add a comment |
$begingroup$
I still have some difficulty to calculate $int frac{1}{u^2+frac{3}{4}}du$. But otherwise very explanatory answer thank you
$endgroup$
– VakiPitsi
Jan 30 at 0:27
1
$begingroup$
Multiply both numerator and denominator by $frac{4}{2}$ to get $frac{4}{3}intfrac{1}{frac{4}{3}u^2+ 1}du$. Now let $v= frac{2}{sqrt{3}}u$ so that $dv= frac{2}{sqrt{3}}du$, $du= frac{sqrt{3}}{2}dv$, and the integral becomes $frac{2}{sqrt{3}}intfrac{1}{v^2+ 1}dv$. That integral is an arctangent.
$endgroup$
– user247327
Jan 30 at 12:57
$begingroup$
I still have some difficulty to calculate $int frac{1}{u^2+frac{3}{4}}du$. But otherwise very explanatory answer thank you
$endgroup$
– VakiPitsi
Jan 30 at 0:27
$begingroup$
I still have some difficulty to calculate $int frac{1}{u^2+frac{3}{4}}du$. But otherwise very explanatory answer thank you
$endgroup$
– VakiPitsi
Jan 30 at 0:27
1
1
$begingroup$
Multiply both numerator and denominator by $frac{4}{2}$ to get $frac{4}{3}intfrac{1}{frac{4}{3}u^2+ 1}du$. Now let $v= frac{2}{sqrt{3}}u$ so that $dv= frac{2}{sqrt{3}}du$, $du= frac{sqrt{3}}{2}dv$, and the integral becomes $frac{2}{sqrt{3}}intfrac{1}{v^2+ 1}dv$. That integral is an arctangent.
$endgroup$
– user247327
Jan 30 at 12:57
$begingroup$
Multiply both numerator and denominator by $frac{4}{2}$ to get $frac{4}{3}intfrac{1}{frac{4}{3}u^2+ 1}du$. Now let $v= frac{2}{sqrt{3}}u$ so that $dv= frac{2}{sqrt{3}}du$, $du= frac{sqrt{3}}{2}dv$, and the integral becomes $frac{2}{sqrt{3}}intfrac{1}{v^2+ 1}dv$. That integral is an arctangent.
$endgroup$
– user247327
Jan 30 at 12:57
add a comment |
$begingroup$
Hint: You can break up your second calculated indefinite integral into two components as so:
$$frac{2-x}{x^2-x+1} = -frac{1}{2}(frac{2x -1}{x^2-x+1}) + frac{3/2}{(x-frac{1}{2})^2 + 3/4}$$
Note that $x^2-x+1 = (x - frac{1}{2})^2 + frac{3}{4}$
answered Jan 29 at 23:44
HyperionHyperion
702111
$endgroup$
add a comment |
$begingroup$
Hint: You can break up your second calculated indefinite integral into two components as so:
$$frac{2-x}{x^2-x+1} = -frac{1}{2}(frac{2x -1}{x^2-x+1}) + frac{3/2}{(x-frac{1}{2})^2 + 3/4}$$
Note that $x^2-x+1 = (x - frac{1}{2})^2 + frac{3}{4}$
answered Jan 29 at 23:44
HyperionHyperion
702111
$endgroup$
add a comment |
$begingroup$
Hint: You can break up your second calculated indefinite integral into two components as so:
$$frac{2-x}{x^2-x+1} = -frac{1}{2}(frac{2x -1}{x^2-x+1}) + frac{3/2}{(x-frac{1}{2})^2 + 3/4}$$
Note that $x^2-x+1 = (x - frac{1}{2})^2 + frac{3}{4}$
answered Jan 29 at 23:44
HyperionHyperion
702111
$endgroup$
Hint: You can break up your second calculated indefinite integral into two components as so:
$$frac{2-x}{x^2-x+1} = -frac{1}{2}(frac{2x -1}{x^2-x+1}) + frac{3/2}{(x-frac{1}{2})^2 + 3/4}$$
Note that $x^2-x+1 = (x - frac{1}{2})^2 + frac{3}{4}$
answered Jan 29 at 23:44
HyperionHyperion
702111
answered Jan 29 at 23:44
HyperionHyperion
702111
answered Jan 29 at 23:44
HyperionHyperion
702111
answered Jan 29 at 23:44
HyperionHyperion
702111
702111
add a comment |
add a comment |
$begingroup$
Alternative approach: Partial fractions.
Recall that for any complex $z$, and any $n=1,2,...$
$$z^{1/n}=|z|^{1/n}expleft[frac{i}{n}(2pi k+arg z)right],qquad k=0,1,...,n-1$$
Then plug in $z=-1$ and $n=3$ to see that $arg z=arg(-1)=pi$ so that in fact,
$$1+x^3=prod_{k=0}^2left(x-expfrac{ipi(2k+1)}3right)$$
So, letting $r_k=expfrac{ipi(2k+1)}3$,
$$frac1{1+x^3}=prod_{k=0}^2frac1{x-r_k}$$
Look! That's a thing we can do partial fractions on! To do so, we say that
$$prod_{k=0}^2frac1{x-r_k}=sum_{k=0}^2frac{b(k)}{x-r_k}$$
$$left(prod_{a=0}^2(x-r_a)right)prod_{k=0}^2frac1{x-r_k}=left(prod_{a=0}^2(x-r_a)right)sum_{k=0}^2frac{b(k)}{x-r_k}$$
$$1=sum_{k=0}^2frac{b(k)}{x-r_k}prod_{a=0}^2(x-r_a)$$
$$1=sum_{k=0}^2b(k)prod_{a=0\ aneq k}^2(x-r_a)$$
then for any $j=0,1,2$, we may plug in $x=r_j$ and notice that all the terms of the sum vanish except for the case $k=j$, which gives
$$1=b(j)prod_{a=0\ aneq j}^2(r_j-r_a)$$
$$b(j)=prod_{a=0\ aneq j}^2frac1{r_j-r_a}$$
Which is an explicit formula for our partial fractions coefficients. Anyway, we may now integrate:
$$I=intfrac{mathrm dx}{1+x^3}=intsum_{k=0}^2frac{b(k)}{x-r_k}mathrm dx$$
$$I=sum_{k=0}^2b(k)intfrac{mathrm dx}{x-r_k}$$
Which is very easily shown to be
$$I=sum_{k=0}^{2}b(k)lnleft|x-r_kright|$$
$$I=b(0)lnleft|x-r_0right|+b(1)lnleft|x-r_1right|+b(2)lnleft|x-r_2right|$$
And since $b(k)=prodlimits_{a=0\ aneq k}^2frac1{r_k-r_a}$, we have that
$$
b(0)=frac1{(r_0-r_1)(r_0-r_2)}\
b(1)=frac1{(r_1-r_0)(r_1-r_2)}\
b(2)=frac1{(r_2-r_0)(r_2-r_1)}$$
And from $exp(itheta)=costheta+isintheta$,
$$
r_0=expfrac{ipi}3=frac{1+isqrt3}2\
r_1=expfrac{3ipi}3=-1\
r_2=expfrac{5ipi}3=frac{1-isqrt3}2
$$
So
$$
b(0)=-frac16(1+isqrt3)\
b(1)=frac16(1+isqrt3)\
b(2)=frac16(-1+isqrt3)
$$
And finally
$$I=-frac{1+isqrt3}6lnleft|x-frac{1+isqrt3}2right|+frac{1+isqrt3}6lnleft|x+1right|+frac{-1+isqrt3}6lnleft|x+frac{-1+isqrt3}2right|+C$$
In fact, using the same method, it can be shown that
$$intfrac{mathrm dx}{1+x^n}=sum_{k=0}^{n-1}lnleft|x-expfrac{ipi(2k+1)}{n}right|prod_{a=0\aneq k}^{n-1}frac1{expfrac{ipi(2k+1)}{n}-expfrac{ipi(2a+1)}{n}}$$
answered Jan 30 at 1:47
clathratusclathratus
5,0641439
$endgroup$
1
$begingroup$
Before I got halfway through reading the solution, I knew it was yours!
$endgroup$
– user150203
Jan 30 at 6:42
$begingroup$
@DavidG Haha thanks. I can usually tell which solutions are yours too :)
$endgroup$
– clathratus
Jan 30 at 21:36
add a comment |
$begingroup$
Alternative approach: Partial fractions.
Recall that for any complex $z$, and any $n=1,2,...$
$$z^{1/n}=|z|^{1/n}expleft[frac{i}{n}(2pi k+arg z)right],qquad k=0,1,...,n-1$$
Then plug in $z=-1$ and $n=3$ to see that $arg z=arg(-1)=pi$ so that in fact,
$$1+x^3=prod_{k=0}^2left(x-expfrac{ipi(2k+1)}3right)$$
So, letting $r_k=expfrac{ipi(2k+1)}3$,
$$frac1{1+x^3}=prod_{k=0}^2frac1{x-r_k}$$
Look! That's a thing we can do partial fractions on! To do so, we say that
$$prod_{k=0}^2frac1{x-r_k}=sum_{k=0}^2frac{b(k)}{x-r_k}$$
$$left(prod_{a=0}^2(x-r_a)right)prod_{k=0}^2frac1{x-r_k}=left(prod_{a=0}^2(x-r_a)right)sum_{k=0}^2frac{b(k)}{x-r_k}$$
$$1=sum_{k=0}^2frac{b(k)}{x-r_k}prod_{a=0}^2(x-r_a)$$
$$1=sum_{k=0}^2b(k)prod_{a=0\ aneq k}^2(x-r_a)$$
then for any $j=0,1,2$, we may plug in $x=r_j$ and notice that all the terms of the sum vanish except for the case $k=j$, which gives
$$1=b(j)prod_{a=0\ aneq j}^2(r_j-r_a)$$
$$b(j)=prod_{a=0\ aneq j}^2frac1{r_j-r_a}$$
Which is an explicit formula for our partial fractions coefficients. Anyway, we may now integrate:
$$I=intfrac{mathrm dx}{1+x^3}=intsum_{k=0}^2frac{b(k)}{x-r_k}mathrm dx$$
$$I=sum_{k=0}^2b(k)intfrac{mathrm dx}{x-r_k}$$
Which is very easily shown to be
$$I=sum_{k=0}^{2}b(k)lnleft|x-r_kright|$$
$$I=b(0)lnleft|x-r_0right|+b(1)lnleft|x-r_1right|+b(2)lnleft|x-r_2right|$$
And since $b(k)=prodlimits_{a=0\ aneq k}^2frac1{r_k-r_a}$, we have that
$$
b(0)=frac1{(r_0-r_1)(r_0-r_2)}\
b(1)=frac1{(r_1-r_0)(r_1-r_2)}\
b(2)=frac1{(r_2-r_0)(r_2-r_1)}$$
And from $exp(itheta)=costheta+isintheta$,
$$
r_0=expfrac{ipi}3=frac{1+isqrt3}2\
r_1=expfrac{3ipi}3=-1\
r_2=expfrac{5ipi}3=frac{1-isqrt3}2
$$
So
$$
b(0)=-frac16(1+isqrt3)\
b(1)=frac16(1+isqrt3)\
b(2)=frac16(-1+isqrt3)
$$
And finally
$$I=-frac{1+isqrt3}6lnleft|x-frac{1+isqrt3}2right|+frac{1+isqrt3}6lnleft|x+1right|+frac{-1+isqrt3}6lnleft|x+frac{-1+isqrt3}2right|+C$$
In fact, using the same method, it can be shown that
$$intfrac{mathrm dx}{1+x^n}=sum_{k=0}^{n-1}lnleft|x-expfrac{ipi(2k+1)}{n}right|prod_{a=0\aneq k}^{n-1}frac1{expfrac{ipi(2k+1)}{n}-expfrac{ipi(2a+1)}{n}}$$
answered Jan 30 at 1:47
clathratusclathratus
5,0641439
$endgroup$
1
$begingroup$
Before I got halfway through reading the solution, I knew it was yours!
$endgroup$
– user150203
Jan 30 at 6:42
$begingroup$
@DavidG Haha thanks. I can usually tell which solutions are yours too :)
$endgroup$
– clathratus
Jan 30 at 21:36
add a comment |
$begingroup$
Alternative approach: Partial fractions.
Recall that for any complex $z$, and any $n=1,2,...$
$$z^{1/n}=|z|^{1/n}expleft[frac{i}{n}(2pi k+arg z)right],qquad k=0,1,...,n-1$$
Then plug in $z=-1$ and $n=3$ to see that $arg z=arg(-1)=pi$ so that in fact,
$$1+x^3=prod_{k=0}^2left(x-expfrac{ipi(2k+1)}3right)$$
So, letting $r_k=expfrac{ipi(2k+1)}3$,
$$frac1{1+x^3}=prod_{k=0}^2frac1{x-r_k}$$
Look! That's a thing we can do partial fractions on! To do so, we say that
$$prod_{k=0}^2frac1{x-r_k}=sum_{k=0}^2frac{b(k)}{x-r_k}$$
$$left(prod_{a=0}^2(x-r_a)right)prod_{k=0}^2frac1{x-r_k}=left(prod_{a=0}^2(x-r_a)right)sum_{k=0}^2frac{b(k)}{x-r_k}$$
$$1=sum_{k=0}^2frac{b(k)}{x-r_k}prod_{a=0}^2(x-r_a)$$
$$1=sum_{k=0}^2b(k)prod_{a=0\ aneq k}^2(x-r_a)$$
then for any $j=0,1,2$, we may plug in $x=r_j$ and notice that all the terms of the sum vanish except for the case $k=j$, which gives
$$1=b(j)prod_{a=0\ aneq j}^2(r_j-r_a)$$
$$b(j)=prod_{a=0\ aneq j}^2frac1{r_j-r_a}$$
Which is an explicit formula for our partial fractions coefficients. Anyway, we may now integrate:
$$I=intfrac{mathrm dx}{1+x^3}=intsum_{k=0}^2frac{b(k)}{x-r_k}mathrm dx$$
$$I=sum_{k=0}^2b(k)intfrac{mathrm dx}{x-r_k}$$
Which is very easily shown to be
$$I=sum_{k=0}^{2}b(k)lnleft|x-r_kright|$$
$$I=b(0)lnleft|x-r_0right|+b(1)lnleft|x-r_1right|+b(2)lnleft|x-r_2right|$$
And since $b(k)=prodlimits_{a=0\ aneq k}^2frac1{r_k-r_a}$, we have that
$$
b(0)=frac1{(r_0-r_1)(r_0-r_2)}\
b(1)=frac1{(r_1-r_0)(r_1-r_2)}\
b(2)=frac1{(r_2-r_0)(r_2-r_1)}$$
And from $exp(itheta)=costheta+isintheta$,
$$
r_0=expfrac{ipi}3=frac{1+isqrt3}2\
r_1=expfrac{3ipi}3=-1\
r_2=expfrac{5ipi}3=frac{1-isqrt3}2
$$
So
$$
b(0)=-frac16(1+isqrt3)\
b(1)=frac16(1+isqrt3)\
b(2)=frac16(-1+isqrt3)
$$
And finally
$$I=-frac{1+isqrt3}6lnleft|x-frac{1+isqrt3}2right|+frac{1+isqrt3}6lnleft|x+1right|+frac{-1+isqrt3}6lnleft|x+frac{-1+isqrt3}2right|+C$$
In fact, using the same method, it can be shown that
$$intfrac{mathrm dx}{1+x^n}=sum_{k=0}^{n-1}lnleft|x-expfrac{ipi(2k+1)}{n}right|prod_{a=0\aneq k}^{n-1}frac1{expfrac{ipi(2k+1)}{n}-expfrac{ipi(2a+1)}{n}}$$
answered Jan 30 at 1:47
clathratusclathratus
5,0641439
$endgroup$
Alternative approach: Partial fractions.
Recall that for any complex $z$, and any $n=1,2,...$
$$z^{1/n}=|z|^{1/n}expleft[frac{i}{n}(2pi k+arg z)right],qquad k=0,1,...,n-1$$
Then plug in $z=-1$ and $n=3$ to see that $arg z=arg(-1)=pi$ so that in fact,
$$1+x^3=prod_{k=0}^2left(x-expfrac{ipi(2k+1)}3right)$$
So, letting $r_k=expfrac{ipi(2k+1)}3$,
$$frac1{1+x^3}=prod_{k=0}^2frac1{x-r_k}$$
Look! That's a thing we can do partial fractions on! To do so, we say that
$$prod_{k=0}^2frac1{x-r_k}=sum_{k=0}^2frac{b(k)}{x-r_k}$$
$$left(prod_{a=0}^2(x-r_a)right)prod_{k=0}^2frac1{x-r_k}=left(prod_{a=0}^2(x-r_a)right)sum_{k=0}^2frac{b(k)}{x-r_k}$$
$$1=sum_{k=0}^2frac{b(k)}{x-r_k}prod_{a=0}^2(x-r_a)$$
$$1=sum_{k=0}^2b(k)prod_{a=0\ aneq k}^2(x-r_a)$$
then for any $j=0,1,2$, we may plug in $x=r_j$ and notice that all the terms of the sum vanish except for the case $k=j$, which gives
$$1=b(j)prod_{a=0\ aneq j}^2(r_j-r_a)$$
$$b(j)=prod_{a=0\ aneq j}^2frac1{r_j-r_a}$$
Which is an explicit formula for our partial fractions coefficients. Anyway, we may now integrate:
$$I=intfrac{mathrm dx}{1+x^3}=intsum_{k=0}^2frac{b(k)}{x-r_k}mathrm dx$$
$$I=sum_{k=0}^2b(k)intfrac{mathrm dx}{x-r_k}$$
Which is very easily shown to be
$$I=sum_{k=0}^{2}b(k)lnleft|x-r_kright|$$
$$I=b(0)lnleft|x-r_0right|+b(1)lnleft|x-r_1right|+b(2)lnleft|x-r_2right|$$
And since $b(k)=prodlimits_{a=0\ aneq k}^2frac1{r_k-r_a}$, we have that
$$
b(0)=frac1{(r_0-r_1)(r_0-r_2)}\
b(1)=frac1{(r_1-r_0)(r_1-r_2)}\
b(2)=frac1{(r_2-r_0)(r_2-r_1)}$$
And from $exp(itheta)=costheta+isintheta$,
$$
r_0=expfrac{ipi}3=frac{1+isqrt3}2\
r_1=expfrac{3ipi}3=-1\
r_2=expfrac{5ipi}3=frac{1-isqrt3}2
$$
So
$$
b(0)=-frac16(1+isqrt3)\
b(1)=frac16(1+isqrt3)\
b(2)=frac16(-1+isqrt3)
$$
And finally
$$I=-frac{1+isqrt3}6lnleft|x-frac{1+isqrt3}2right|+frac{1+isqrt3}6lnleft|x+1right|+frac{-1+isqrt3}6lnleft|x+frac{-1+isqrt3}2right|+C$$
In fact, using the same method, it can be shown that
$$intfrac{mathrm dx}{1+x^n}=sum_{k=0}^{n-1}lnleft|x-expfrac{ipi(2k+1)}{n}right|prod_{a=0\aneq k}^{n-1}frac1{expfrac{ipi(2k+1)}{n}-expfrac{ipi(2a+1)}{n}}$$
answered Jan 30 at 1:47
clathratusclathratus
5,0641439
edited Jan 30 at 1:56
answered Jan 30 at 1:47
clathratusclathratus
5,0641439
answered Jan 30 at 1:47
clathratusclathratus
5,0641439
answered Jan 30 at 1:47
clathratusclathratus
5,0641439
5,0641439
1
$begingroup$
Before I got halfway through reading the solution, I knew it was yours!
$endgroup$
– user150203
Jan 30 at 6:42
$begingroup$
@DavidG Haha thanks. I can usually tell which solutions are yours too :)
$endgroup$
– clathratus
Jan 30 at 21:36
add a comment |
1
$begingroup$
Before I got halfway through reading the solution, I knew it was yours!
$endgroup$
– user150203
Jan 30 at 6:42
$begingroup$
@DavidG Haha thanks. I can usually tell which solutions are yours too :)
$endgroup$
– clathratus
Jan 30 at 21:36
1
1
$begingroup$
Before I got halfway through reading the solution, I knew it was yours!
$endgroup$
– user150203
Jan 30 at 6:42
$begingroup$
Before I got halfway through reading the solution, I knew it was yours!
$endgroup$
– user150203
Jan 30 at 6:42
$begingroup$
@DavidG Haha thanks. I can usually tell which solutions are yours too :)
$endgroup$
– clathratus
Jan 30 at 21:36
$begingroup$
@DavidG Haha thanks. I can usually tell which solutions are yours too :)
$endgroup$
– clathratus
Jan 30 at 21:36
add a comment |
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