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Calculating $int frac{dx}{(x^2+1)^3}$ integral
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Would anyone help me calculate the following integral? $int frac{dx}{(x^2+1)^3}$
During our lecutre we've done very similiar one, $int frac{dx}{(x^2+1)^2}$ like that:
$int frac{dx}{(x^2+1)^2} = int frac{x^2+1-x^2}{(x^2+1)^2}dx = int frac{1}{x^2+1}dx - int frac{x^2}{(x^2+1)^2}dx = $
$= Biggrrvert begin{equation}
begin{split}
& u = x quad v' =frac{x}{(x^2+1)^2} =frac{1(x^2+1)'}{2(x^2+1)^2}\
& u' = 1 quad v = -frac{1}{2} frac{1}{x^2+1}
end{split}
end{equation} Biggrrvert$
$= arctan x - (-xfrac{1}{2}frac{1}{x^2+1} + frac{1}{2} int frac{dx}{x^2+1})$
$= arctan x + frac{x}{2(x^2+1)} - frac{1}{2}arctan x + C = frac{1}{2}arctan x + frac{x}{2(x^2+1)} + C$
Thank you.
real-analysis calculus integration
asked Jan 9 at 16:18
wenoweno
30511
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add a comment |
$begingroup$
Would anyone help me calculate the following integral? $int frac{dx}{(x^2+1)^3}$
During our lecutre we've done very similiar one, $int frac{dx}{(x^2+1)^2}$ like that:
$int frac{dx}{(x^2+1)^2} = int frac{x^2+1-x^2}{(x^2+1)^2}dx = int frac{1}{x^2+1}dx - int frac{x^2}{(x^2+1)^2}dx = $
$= Biggrrvert begin{equation}
begin{split}
& u = x quad v' =frac{x}{(x^2+1)^2} =frac{1(x^2+1)'}{2(x^2+1)^2}\
& u' = 1 quad v = -frac{1}{2} frac{1}{x^2+1}
end{split}
end{equation} Biggrrvert$
$= arctan x - (-xfrac{1}{2}frac{1}{x^2+1} + frac{1}{2} int frac{dx}{x^2+1})$
$= arctan x + frac{x}{2(x^2+1)} - frac{1}{2}arctan x + C = frac{1}{2}arctan x + frac{x}{2(x^2+1)} + C$
Thank you.
real-analysis calculus integration
asked Jan 9 at 16:18
wenoweno
30511
$endgroup$
add a comment |
$begingroup$
Would anyone help me calculate the following integral? $int frac{dx}{(x^2+1)^3}$
During our lecutre we've done very similiar one, $int frac{dx}{(x^2+1)^2}$ like that:
$int frac{dx}{(x^2+1)^2} = int frac{x^2+1-x^2}{(x^2+1)^2}dx = int frac{1}{x^2+1}dx - int frac{x^2}{(x^2+1)^2}dx = $
$= Biggrrvert begin{equation}
begin{split}
& u = x quad v' =frac{x}{(x^2+1)^2} =frac{1(x^2+1)'}{2(x^2+1)^2}\
& u' = 1 quad v = -frac{1}{2} frac{1}{x^2+1}
end{split}
end{equation} Biggrrvert$
$= arctan x - (-xfrac{1}{2}frac{1}{x^2+1} + frac{1}{2} int frac{dx}{x^2+1})$
$= arctan x + frac{x}{2(x^2+1)} - frac{1}{2}arctan x + C = frac{1}{2}arctan x + frac{x}{2(x^2+1)} + C$
Thank you.
real-analysis calculus integration
asked Jan 9 at 16:18
wenoweno
30511
$endgroup$
Would anyone help me calculate the following integral? $int frac{dx}{(x^2+1)^3}$
During our lecutre we've done very similiar one, $int frac{dx}{(x^2+1)^2}$ like that:
$int frac{dx}{(x^2+1)^2} = int frac{x^2+1-x^2}{(x^2+1)^2}dx = int frac{1}{x^2+1}dx - int frac{x^2}{(x^2+1)^2}dx = $
$= Biggrrvert begin{equation}
begin{split}
& u = x quad v' =frac{x}{(x^2+1)^2} =frac{1(x^2+1)'}{2(x^2+1)^2}\
& u' = 1 quad v = -frac{1}{2} frac{1}{x^2+1}
end{split}
end{equation} Biggrrvert$
$= arctan x - (-xfrac{1}{2}frac{1}{x^2+1} + frac{1}{2} int frac{dx}{x^2+1})$
$= arctan x + frac{x}{2(x^2+1)} - frac{1}{2}arctan x + C = frac{1}{2}arctan x + frac{x}{2(x^2+1)} + C$
Thank you.
real-analysis calculus integration
real-analysis calculus integration
asked Jan 9 at 16:18
wenoweno
30511
asked Jan 9 at 16:18
wenoweno
30511
asked Jan 9 at 16:18
wenoweno
30511
asked Jan 9 at 16:18
wenoweno
30511
asked Jan 9 at 16:18
wenoweno
30511
30511
add a comment |
add a comment |
4 Answers
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We will find a general reduction formula for the integral
$$I_n=intfrac{dx}{(ax^2+b)^n}$$
Integration by parts with
$$dv=dxRightarrow v=x\ u=frac1{(ax^2+b)^n}Rightarrow du=frac{-2anx}{(ax^2+b)^{n+1}}dx$$
Yields
$$I_n=frac{x}{(ax^2+b)^n}+2nintfrac{ax^2}{(ax^2+b)^{n+1}}dx$$
$$I_n=frac{x}{(ax^2+b)^n}+2nintfrac{ax^2+b}{(ax^2+b)^{n+1}}dx-2bnintfrac{dx}{(ax^2+b)^{n+1}}$$
$$I_n=frac{x}{(ax^2+b)^n}+2nintfrac{dx}{(ax^2+b)^{n}}-2bnI_{n+1}$$
$$I_n=frac{x}{(ax^2+b)^n}+2nI_n-2bnI_{n+1}$$
$$2bnI_{n+1}=frac{x}{(ax^2+b)^n}+(2n-1)I_n$$
$$I_{n+1}=frac{x}{2bn(ax^2+b)^n}+frac{2n-1}{2bn}I_n$$
replacing $n+1$ with $n$,
$$I_{n}=frac{x}{2b(n-1)(ax^2+b)^{n-1}}+frac{2n-3}{2b(n-1)}I_{n-1}$$
Now for the base case $I_1$:
$$I_1=intfrac{dx}{ax^2+b}$$
Let $$x=sqrt{frac{b}a}tan uRightarrow dx=sqrt{frac{b}a}sec^2u, du$$
So $$I_1=sqrt{frac{b}a}intfrac{sec^2u}{btan^2u+b}du$$
$$I_1=frac1{sqrt{ab}}intfrac{sec^2u}{sec^2u}du$$
$$I_1=frac1{sqrt{ab}}int du$$
$$I_1=frac{u}{sqrt{ab}}$$
$$I_1=frac1{sqrt{ab}}arctansqrt{frac{a}{b}}x+C$$
Plug in your specific $a,b$ and $n$, and you're good to go.
Edit: Whatever I'll just give you the answer.
Plugging in $a=1, b=1, n=3$ we have
$$I_{3}=frac{x}{4(x^2+1)^{2}}+frac{3}{4}I_{2}$$
Note that
$$I_{2}=frac{x}{2(x^2+1)}+frac{1}{2}I_{1}$$
$$I_{2}=frac{x}{2(x^2+1)}+frac{1}{2}arctan x+C$$
So
$$I_{3}=frac{x}{4(x^2+1)^{2}}+frac{3x}{8(x^2+1)}+frac{3}{8}arctan x+C$$
answered Jan 9 at 16:58
clathratusclathratus
4,174336
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So why not use the same trick? Since you know from lecture what $$f_2(x) = int frac{dx}{left(x^2+1right)^2}$$ is, you now have
$$
begin{split}
f_3(x)
&= int frac{dx}{left(x^2+1right)^3}\
&= int frac{dx}{left(x^2+1right)^2} - int x cdot frac{xdx}{left(x^2+1right)^3}\
&= f_2(x) - text{proceed by parts as before}
end{split}
$$
answered Jan 9 at 16:24
gt6989bgt6989b
33.9k22455
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Use $$int frac{dx}{(x^2+1)^3}=frac{x}{4(x^2+1)^2}+frac{3}{4}int frac{dx}{(x^2+1)^2}$$ and then use the value for the integral you already calculated.
answered Jan 9 at 16:23
aledenaleden
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I used:
$$p=x^2+1to dx =frac{1}{2sqrt{p-1}}$$
which leads to:
$$int{frac{dx}{(x^2+1)^3}=frac12 int{frac{dp}{psqrt{p-1}}}}$$
Then the substitution $$q=sqrt{p-1}to dp=2qspace dq$$
Leads to $$frac12intfrac{dp}{psqrt{p-1}}=frac12intfrac{2q}{q(q^2+1)}dq=intfrac{dq}{q^2+1}$$
Which is resolved simply.
answered Jan 9 at 16:34
Rhys HughesRhys Hughes
5,8531529
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4 Answers
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active
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votes
4 Answers
4
active
oldest
votes
active
oldest
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active
oldest
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$begingroup$
We will find a general reduction formula for the integral
$$I_n=intfrac{dx}{(ax^2+b)^n}$$
Integration by parts with
$$dv=dxRightarrow v=x\ u=frac1{(ax^2+b)^n}Rightarrow du=frac{-2anx}{(ax^2+b)^{n+1}}dx$$
Yields
$$I_n=frac{x}{(ax^2+b)^n}+2nintfrac{ax^2}{(ax^2+b)^{n+1}}dx$$
$$I_n=frac{x}{(ax^2+b)^n}+2nintfrac{ax^2+b}{(ax^2+b)^{n+1}}dx-2bnintfrac{dx}{(ax^2+b)^{n+1}}$$
$$I_n=frac{x}{(ax^2+b)^n}+2nintfrac{dx}{(ax^2+b)^{n}}-2bnI_{n+1}$$
$$I_n=frac{x}{(ax^2+b)^n}+2nI_n-2bnI_{n+1}$$
$$2bnI_{n+1}=frac{x}{(ax^2+b)^n}+(2n-1)I_n$$
$$I_{n+1}=frac{x}{2bn(ax^2+b)^n}+frac{2n-1}{2bn}I_n$$
replacing $n+1$ with $n$,
$$I_{n}=frac{x}{2b(n-1)(ax^2+b)^{n-1}}+frac{2n-3}{2b(n-1)}I_{n-1}$$
Now for the base case $I_1$:
$$I_1=intfrac{dx}{ax^2+b}$$
Let $$x=sqrt{frac{b}a}tan uRightarrow dx=sqrt{frac{b}a}sec^2u, du$$
So $$I_1=sqrt{frac{b}a}intfrac{sec^2u}{btan^2u+b}du$$
$$I_1=frac1{sqrt{ab}}intfrac{sec^2u}{sec^2u}du$$
$$I_1=frac1{sqrt{ab}}int du$$
$$I_1=frac{u}{sqrt{ab}}$$
$$I_1=frac1{sqrt{ab}}arctansqrt{frac{a}{b}}x+C$$
Plug in your specific $a,b$ and $n$, and you're good to go.
Edit: Whatever I'll just give you the answer.
Plugging in $a=1, b=1, n=3$ we have
$$I_{3}=frac{x}{4(x^2+1)^{2}}+frac{3}{4}I_{2}$$
Note that
$$I_{2}=frac{x}{2(x^2+1)}+frac{1}{2}I_{1}$$
$$I_{2}=frac{x}{2(x^2+1)}+frac{1}{2}arctan x+C$$
So
$$I_{3}=frac{x}{4(x^2+1)^{2}}+frac{3x}{8(x^2+1)}+frac{3}{8}arctan x+C$$
answered Jan 9 at 16:58
clathratusclathratus
4,174336
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add a comment |
$begingroup$
We will find a general reduction formula for the integral
$$I_n=intfrac{dx}{(ax^2+b)^n}$$
Integration by parts with
$$dv=dxRightarrow v=x\ u=frac1{(ax^2+b)^n}Rightarrow du=frac{-2anx}{(ax^2+b)^{n+1}}dx$$
Yields
$$I_n=frac{x}{(ax^2+b)^n}+2nintfrac{ax^2}{(ax^2+b)^{n+1}}dx$$
$$I_n=frac{x}{(ax^2+b)^n}+2nintfrac{ax^2+b}{(ax^2+b)^{n+1}}dx-2bnintfrac{dx}{(ax^2+b)^{n+1}}$$
$$I_n=frac{x}{(ax^2+b)^n}+2nintfrac{dx}{(ax^2+b)^{n}}-2bnI_{n+1}$$
$$I_n=frac{x}{(ax^2+b)^n}+2nI_n-2bnI_{n+1}$$
$$2bnI_{n+1}=frac{x}{(ax^2+b)^n}+(2n-1)I_n$$
$$I_{n+1}=frac{x}{2bn(ax^2+b)^n}+frac{2n-1}{2bn}I_n$$
replacing $n+1$ with $n$,
$$I_{n}=frac{x}{2b(n-1)(ax^2+b)^{n-1}}+frac{2n-3}{2b(n-1)}I_{n-1}$$
Now for the base case $I_1$:
$$I_1=intfrac{dx}{ax^2+b}$$
Let $$x=sqrt{frac{b}a}tan uRightarrow dx=sqrt{frac{b}a}sec^2u, du$$
So $$I_1=sqrt{frac{b}a}intfrac{sec^2u}{btan^2u+b}du$$
$$I_1=frac1{sqrt{ab}}intfrac{sec^2u}{sec^2u}du$$
$$I_1=frac1{sqrt{ab}}int du$$
$$I_1=frac{u}{sqrt{ab}}$$
$$I_1=frac1{sqrt{ab}}arctansqrt{frac{a}{b}}x+C$$
Plug in your specific $a,b$ and $n$, and you're good to go.
Edit: Whatever I'll just give you the answer.
Plugging in $a=1, b=1, n=3$ we have
$$I_{3}=frac{x}{4(x^2+1)^{2}}+frac{3}{4}I_{2}$$
Note that
$$I_{2}=frac{x}{2(x^2+1)}+frac{1}{2}I_{1}$$
$$I_{2}=frac{x}{2(x^2+1)}+frac{1}{2}arctan x+C$$
So
$$I_{3}=frac{x}{4(x^2+1)^{2}}+frac{3x}{8(x^2+1)}+frac{3}{8}arctan x+C$$
answered Jan 9 at 16:58
clathratusclathratus
4,174336
$endgroup$
add a comment |
$begingroup$
We will find a general reduction formula for the integral
$$I_n=intfrac{dx}{(ax^2+b)^n}$$
Integration by parts with
$$dv=dxRightarrow v=x\ u=frac1{(ax^2+b)^n}Rightarrow du=frac{-2anx}{(ax^2+b)^{n+1}}dx$$
Yields
$$I_n=frac{x}{(ax^2+b)^n}+2nintfrac{ax^2}{(ax^2+b)^{n+1}}dx$$
$$I_n=frac{x}{(ax^2+b)^n}+2nintfrac{ax^2+b}{(ax^2+b)^{n+1}}dx-2bnintfrac{dx}{(ax^2+b)^{n+1}}$$
$$I_n=frac{x}{(ax^2+b)^n}+2nintfrac{dx}{(ax^2+b)^{n}}-2bnI_{n+1}$$
$$I_n=frac{x}{(ax^2+b)^n}+2nI_n-2bnI_{n+1}$$
$$2bnI_{n+1}=frac{x}{(ax^2+b)^n}+(2n-1)I_n$$
$$I_{n+1}=frac{x}{2bn(ax^2+b)^n}+frac{2n-1}{2bn}I_n$$
replacing $n+1$ with $n$,
$$I_{n}=frac{x}{2b(n-1)(ax^2+b)^{n-1}}+frac{2n-3}{2b(n-1)}I_{n-1}$$
Now for the base case $I_1$:
$$I_1=intfrac{dx}{ax^2+b}$$
Let $$x=sqrt{frac{b}a}tan uRightarrow dx=sqrt{frac{b}a}sec^2u, du$$
So $$I_1=sqrt{frac{b}a}intfrac{sec^2u}{btan^2u+b}du$$
$$I_1=frac1{sqrt{ab}}intfrac{sec^2u}{sec^2u}du$$
$$I_1=frac1{sqrt{ab}}int du$$
$$I_1=frac{u}{sqrt{ab}}$$
$$I_1=frac1{sqrt{ab}}arctansqrt{frac{a}{b}}x+C$$
Plug in your specific $a,b$ and $n$, and you're good to go.
Edit: Whatever I'll just give you the answer.
Plugging in $a=1, b=1, n=3$ we have
$$I_{3}=frac{x}{4(x^2+1)^{2}}+frac{3}{4}I_{2}$$
Note that
$$I_{2}=frac{x}{2(x^2+1)}+frac{1}{2}I_{1}$$
$$I_{2}=frac{x}{2(x^2+1)}+frac{1}{2}arctan x+C$$
So
$$I_{3}=frac{x}{4(x^2+1)^{2}}+frac{3x}{8(x^2+1)}+frac{3}{8}arctan x+C$$
answered Jan 9 at 16:58
clathratusclathratus
4,174336
$endgroup$
We will find a general reduction formula for the integral
$$I_n=intfrac{dx}{(ax^2+b)^n}$$
Integration by parts with
$$dv=dxRightarrow v=x\ u=frac1{(ax^2+b)^n}Rightarrow du=frac{-2anx}{(ax^2+b)^{n+1}}dx$$
Yields
$$I_n=frac{x}{(ax^2+b)^n}+2nintfrac{ax^2}{(ax^2+b)^{n+1}}dx$$
$$I_n=frac{x}{(ax^2+b)^n}+2nintfrac{ax^2+b}{(ax^2+b)^{n+1}}dx-2bnintfrac{dx}{(ax^2+b)^{n+1}}$$
$$I_n=frac{x}{(ax^2+b)^n}+2nintfrac{dx}{(ax^2+b)^{n}}-2bnI_{n+1}$$
$$I_n=frac{x}{(ax^2+b)^n}+2nI_n-2bnI_{n+1}$$
$$2bnI_{n+1}=frac{x}{(ax^2+b)^n}+(2n-1)I_n$$
$$I_{n+1}=frac{x}{2bn(ax^2+b)^n}+frac{2n-1}{2bn}I_n$$
replacing $n+1$ with $n$,
$$I_{n}=frac{x}{2b(n-1)(ax^2+b)^{n-1}}+frac{2n-3}{2b(n-1)}I_{n-1}$$
Now for the base case $I_1$:
$$I_1=intfrac{dx}{ax^2+b}$$
Let $$x=sqrt{frac{b}a}tan uRightarrow dx=sqrt{frac{b}a}sec^2u, du$$
So $$I_1=sqrt{frac{b}a}intfrac{sec^2u}{btan^2u+b}du$$
$$I_1=frac1{sqrt{ab}}intfrac{sec^2u}{sec^2u}du$$
$$I_1=frac1{sqrt{ab}}int du$$
$$I_1=frac{u}{sqrt{ab}}$$
$$I_1=frac1{sqrt{ab}}arctansqrt{frac{a}{b}}x+C$$
Plug in your specific $a,b$ and $n$, and you're good to go.
Edit: Whatever I'll just give you the answer.
Plugging in $a=1, b=1, n=3$ we have
$$I_{3}=frac{x}{4(x^2+1)^{2}}+frac{3}{4}I_{2}$$
Note that
$$I_{2}=frac{x}{2(x^2+1)}+frac{1}{2}I_{1}$$
$$I_{2}=frac{x}{2(x^2+1)}+frac{1}{2}arctan x+C$$
So
$$I_{3}=frac{x}{4(x^2+1)^{2}}+frac{3x}{8(x^2+1)}+frac{3}{8}arctan x+C$$
answered Jan 9 at 16:58
clathratusclathratus
4,174336
edited Jan 9 at 23:27
answered Jan 9 at 16:58
clathratusclathratus
4,174336
answered Jan 9 at 16:58
clathratusclathratus
4,174336
answered Jan 9 at 16:58
clathratusclathratus
4,174336
4,174336
add a comment |
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$begingroup$
So why not use the same trick? Since you know from lecture what $$f_2(x) = int frac{dx}{left(x^2+1right)^2}$$ is, you now have
$$
begin{split}
f_3(x)
&= int frac{dx}{left(x^2+1right)^3}\
&= int frac{dx}{left(x^2+1right)^2} - int x cdot frac{xdx}{left(x^2+1right)^3}\
&= f_2(x) - text{proceed by parts as before}
end{split}
$$
answered Jan 9 at 16:24
gt6989bgt6989b
33.9k22455
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add a comment |
$begingroup$
So why not use the same trick? Since you know from lecture what $$f_2(x) = int frac{dx}{left(x^2+1right)^2}$$ is, you now have
$$
begin{split}
f_3(x)
&= int frac{dx}{left(x^2+1right)^3}\
&= int frac{dx}{left(x^2+1right)^2} - int x cdot frac{xdx}{left(x^2+1right)^3}\
&= f_2(x) - text{proceed by parts as before}
end{split}
$$
answered Jan 9 at 16:24
gt6989bgt6989b
33.9k22455
$endgroup$
add a comment |
$begingroup$
So why not use the same trick? Since you know from lecture what $$f_2(x) = int frac{dx}{left(x^2+1right)^2}$$ is, you now have
$$
begin{split}
f_3(x)
&= int frac{dx}{left(x^2+1right)^3}\
&= int frac{dx}{left(x^2+1right)^2} - int x cdot frac{xdx}{left(x^2+1right)^3}\
&= f_2(x) - text{proceed by parts as before}
end{split}
$$
answered Jan 9 at 16:24
gt6989bgt6989b
33.9k22455
$endgroup$
So why not use the same trick? Since you know from lecture what $$f_2(x) = int frac{dx}{left(x^2+1right)^2}$$ is, you now have
$$
begin{split}
f_3(x)
&= int frac{dx}{left(x^2+1right)^3}\
&= int frac{dx}{left(x^2+1right)^2} - int x cdot frac{xdx}{left(x^2+1right)^3}\
&= f_2(x) - text{proceed by parts as before}
end{split}
$$
answered Jan 9 at 16:24
gt6989bgt6989b
33.9k22455
answered Jan 9 at 16:24
gt6989bgt6989b
33.9k22455
answered Jan 9 at 16:24
gt6989bgt6989b
33.9k22455
answered Jan 9 at 16:24
gt6989bgt6989b
33.9k22455
33.9k22455
add a comment |
add a comment |
$begingroup$
Use $$int frac{dx}{(x^2+1)^3}=frac{x}{4(x^2+1)^2}+frac{3}{4}int frac{dx}{(x^2+1)^2}$$ and then use the value for the integral you already calculated.
answered Jan 9 at 16:23
aledenaleden
2,057511
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$begingroup$
Use $$int frac{dx}{(x^2+1)^3}=frac{x}{4(x^2+1)^2}+frac{3}{4}int frac{dx}{(x^2+1)^2}$$ and then use the value for the integral you already calculated.
answered Jan 9 at 16:23
aledenaleden
2,057511
$endgroup$
add a comment |
$begingroup$
Use $$int frac{dx}{(x^2+1)^3}=frac{x}{4(x^2+1)^2}+frac{3}{4}int frac{dx}{(x^2+1)^2}$$ and then use the value for the integral you already calculated.
answered Jan 9 at 16:23
aledenaleden
2,057511
$endgroup$
Use $$int frac{dx}{(x^2+1)^3}=frac{x}{4(x^2+1)^2}+frac{3}{4}int frac{dx}{(x^2+1)^2}$$ and then use the value for the integral you already calculated.
answered Jan 9 at 16:23
aledenaleden
2,057511
answered Jan 9 at 16:23
aledenaleden
2,057511
answered Jan 9 at 16:23
aledenaleden
2,057511
answered Jan 9 at 16:23
aledenaleden
2,057511
2,057511
add a comment |
add a comment |
$begingroup$
I used:
$$p=x^2+1to dx =frac{1}{2sqrt{p-1}}$$
which leads to:
$$int{frac{dx}{(x^2+1)^3}=frac12 int{frac{dp}{psqrt{p-1}}}}$$
Then the substitution $$q=sqrt{p-1}to dp=2qspace dq$$
Leads to $$frac12intfrac{dp}{psqrt{p-1}}=frac12intfrac{2q}{q(q^2+1)}dq=intfrac{dq}{q^2+1}$$
Which is resolved simply.
answered Jan 9 at 16:34
Rhys HughesRhys Hughes
5,8531529
$endgroup$
add a comment |
$begingroup$
I used:
$$p=x^2+1to dx =frac{1}{2sqrt{p-1}}$$
which leads to:
$$int{frac{dx}{(x^2+1)^3}=frac12 int{frac{dp}{psqrt{p-1}}}}$$
Then the substitution $$q=sqrt{p-1}to dp=2qspace dq$$
Leads to $$frac12intfrac{dp}{psqrt{p-1}}=frac12intfrac{2q}{q(q^2+1)}dq=intfrac{dq}{q^2+1}$$
Which is resolved simply.
answered Jan 9 at 16:34
Rhys HughesRhys Hughes
5,8531529
$endgroup$
add a comment |
$begingroup$
I used:
$$p=x^2+1to dx =frac{1}{2sqrt{p-1}}$$
which leads to:
$$int{frac{dx}{(x^2+1)^3}=frac12 int{frac{dp}{psqrt{p-1}}}}$$
Then the substitution $$q=sqrt{p-1}to dp=2qspace dq$$
Leads to $$frac12intfrac{dp}{psqrt{p-1}}=frac12intfrac{2q}{q(q^2+1)}dq=intfrac{dq}{q^2+1}$$
Which is resolved simply.
answered Jan 9 at 16:34
Rhys HughesRhys Hughes
5,8531529
$endgroup$
I used:
$$p=x^2+1to dx =frac{1}{2sqrt{p-1}}$$
which leads to:
$$int{frac{dx}{(x^2+1)^3}=frac12 int{frac{dp}{psqrt{p-1}}}}$$
Then the substitution $$q=sqrt{p-1}to dp=2qspace dq$$
Leads to $$frac12intfrac{dp}{psqrt{p-1}}=frac12intfrac{2q}{q(q^2+1)}dq=intfrac{dq}{q^2+1}$$
Which is resolved simply.
answered Jan 9 at 16:34
Rhys HughesRhys Hughes
5,8531529
answered Jan 9 at 16:34
Rhys HughesRhys Hughes
5,8531529
answered Jan 9 at 16:34
Rhys HughesRhys Hughes
5,8531529
answered Jan 9 at 16:34
Rhys HughesRhys Hughes
5,8531529
5,8531529
add a comment |
add a comment |
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