Mixing
Evaluate the integral $int frac{x^2 + 1}{x^4 + x^2 +1} dx$
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1
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Evaluate the integral $int frac{x^2 + 1}{x^4 + x^2 +1} dx$
The book hint is:
Take $u = x- (1/x)$ but still I do not understand why this hint works, could anyone explain to me the intuition behind this hint?
The book answer is:
calculus real-analysis integration analysis
asked 18 hours ago
hopefully
2079
add a comment |
up vote
1
down vote
favorite
Evaluate the integral $int frac{x^2 + 1}{x^4 + x^2 +1} dx$
The book hint is:
Take $u = x- (1/x)$ but still I do not understand why this hint works, could anyone explain to me the intuition behind this hint?
The book answer is:
calculus real-analysis integration analysis
asked 18 hours ago
hopefully
2079
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Evaluate the integral $int frac{x^2 + 1}{x^4 + x^2 +1} dx$
The book hint is:
Take $u = x- (1/x)$ but still I do not understand why this hint works, could anyone explain to me the intuition behind this hint?
The book answer is:
calculus real-analysis integration analysis
asked 18 hours ago
hopefully
2079
Evaluate the integral $int frac{x^2 + 1}{x^4 + x^2 +1} dx$
The book hint is:
Take $u = x- (1/x)$ but still I do not understand why this hint works, could anyone explain to me the intuition behind this hint?
The book answer is:
calculus real-analysis integration analysis
calculus real-analysis integration analysis
asked 18 hours ago
hopefully
2079
asked 18 hours ago
hopefully
2079
edited 17 hours ago
asked 18 hours ago
hopefully
2079
asked 18 hours ago
hopefully
2079
asked 18 hours ago
hopefully
2079
2079
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
- Take $x^2$ common out from numerator and denominator. Then your fraction becomes $$frac{1+frac{1}{x^2}}{x^{2}+frac{1}{x^2}+1}$$ Now write $x^{2}+frac{1}{x^2} = (x-frac{1}{x})^{2}+2$
answered 18 hours ago
crskhr
3,740925
But I can see that the final answer in the book contains 2 arctan ...... could you please explain this also for me?
– hopefully
17 hours ago
Sure. After substitution, $t=x-frac{1}{x}$, the integral becomes $int frac{1}{1+t^2} , dt$ and integral of this is given by $arctan(t)$
– crskhr
17 hours ago
No I do not mean this
– hopefully
17 hours ago
My answer is (1/3) arctan (x^2 -1)/sqrt {3} x ..... but the book answer is
– hopefully
17 hours ago
I will include it in the main question above
– hopefully
17 hours ago
|
show 1 more comment
up vote
0
down vote
It turns out that we must compute the integral in the following way:
$$I=intfrac{x^2+1}{x^4+x^2+1}mathrm{d}x$$
$$I=intfrac{x^2+1}{(x^2-x+1)(x^2+x+1)}mathrm{d}x$$
After a fraction decomposition,
$$I=frac12intfrac{mathrm{d}x}{x^2+x+1}+frac12intfrac{mathrm{d}x}{x^2-x+1}$$
Now we focus on
$$I_1=intfrac{mathrm{d}x}{x^2+x+1}$$
Completing the square in the denominator produces
$$I_1=intfrac{mathrm{d}x}{(x+frac12)^2+frac34}$$
Then the substitution $u=frac1{sqrt{3}}(2x+1)$ gives
$$I_1=frac2{sqrt{3}}intfrac{mathrm{d}u}{u^2+1}$$
$$I_1=frac2{sqrt{3}}arctanfrac{2x+1}{sqrt{3}}$$
Similarly,
$$I_2=intfrac{mathrm{d}x}{x^2-x+1}$$
$$I_2=intfrac{mathrm{d}x}{(x-frac12)^2+frac34}$$
And the substitution $u=x-frac12$ carries us to
$$I_2=frac2{sqrt{3}}arctanfrac{2x-1}{sqrt{3}}$$
Plugging in:
$$I=frac1{sqrt{3}}bigg(arctanfrac{2x+1}{sqrt{3}}+arctanfrac{2x-1}{sqrt{3}}bigg)+C$$
Which is confusing, because nowhere did we use the suggested substitution. Furthermore, using the suggested substitution produces
$$I=intfrac{mathrm{d}u}{u^2+3}$$
Which is easily shown to be
$$I=frac1{sqrt{3}}arctanfrac{x^2+1}{xsqrt{3}}$$
But apparently that is incorrect. I am really unsure how this is the case. Try asking your teacher.
answered 4 hours ago
clathratus
1,729219
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
- Take $x^2$ common out from numerator and denominator. Then your fraction becomes $$frac{1+frac{1}{x^2}}{x^{2}+frac{1}{x^2}+1}$$ Now write $x^{2}+frac{1}{x^2} = (x-frac{1}{x})^{2}+2$
answered 18 hours ago
crskhr
3,740925
But I can see that the final answer in the book contains 2 arctan ...... could you please explain this also for me?
– hopefully
17 hours ago
Sure. After substitution, $t=x-frac{1}{x}$, the integral becomes $int frac{1}{1+t^2} , dt$ and integral of this is given by $arctan(t)$
– crskhr
17 hours ago
No I do not mean this
– hopefully
17 hours ago
My answer is (1/3) arctan (x^2 -1)/sqrt {3} x ..... but the book answer is
– hopefully
17 hours ago
I will include it in the main question above
– hopefully
17 hours ago
|
show 1 more comment
up vote
2
down vote
accepted
- Take $x^2$ common out from numerator and denominator. Then your fraction becomes $$frac{1+frac{1}{x^2}}{x^{2}+frac{1}{x^2}+1}$$ Now write $x^{2}+frac{1}{x^2} = (x-frac{1}{x})^{2}+2$
answered 18 hours ago
crskhr
3,740925
But I can see that the final answer in the book contains 2 arctan ...... could you please explain this also for me?
– hopefully
17 hours ago
Sure. After substitution, $t=x-frac{1}{x}$, the integral becomes $int frac{1}{1+t^2} , dt$ and integral of this is given by $arctan(t)$
– crskhr
17 hours ago
No I do not mean this
– hopefully
17 hours ago
My answer is (1/3) arctan (x^2 -1)/sqrt {3} x ..... but the book answer is
– hopefully
17 hours ago
I will include it in the main question above
– hopefully
17 hours ago
|
show 1 more comment
up vote
2
down vote
accepted
up vote
2
down vote
accepted
- Take $x^2$ common out from numerator and denominator. Then your fraction becomes $$frac{1+frac{1}{x^2}}{x^{2}+frac{1}{x^2}+1}$$ Now write $x^{2}+frac{1}{x^2} = (x-frac{1}{x})^{2}+2$
answered 18 hours ago
crskhr
3,740925
- Take $x^2$ common out from numerator and denominator. Then your fraction becomes $$frac{1+frac{1}{x^2}}{x^{2}+frac{1}{x^2}+1}$$ Now write $x^{2}+frac{1}{x^2} = (x-frac{1}{x})^{2}+2$
answered 18 hours ago
crskhr
3,740925
answered 18 hours ago
crskhr
3,740925
answered 18 hours ago
crskhr
3,740925
answered 18 hours ago
crskhr
3,740925
3,740925
But I can see that the final answer in the book contains 2 arctan ...... could you please explain this also for me?
– hopefully
17 hours ago
Sure. After substitution, $t=x-frac{1}{x}$, the integral becomes $int frac{1}{1+t^2} , dt$ and integral of this is given by $arctan(t)$
– crskhr
17 hours ago
No I do not mean this
– hopefully
17 hours ago
My answer is (1/3) arctan (x^2 -1)/sqrt {3} x ..... but the book answer is
– hopefully
17 hours ago
I will include it in the main question above
– hopefully
17 hours ago
|
show 1 more comment
But I can see that the final answer in the book contains 2 arctan ...... could you please explain this also for me?
– hopefully
17 hours ago
Sure. After substitution, $t=x-frac{1}{x}$, the integral becomes $int frac{1}{1+t^2} , dt$ and integral of this is given by $arctan(t)$
– crskhr
17 hours ago
No I do not mean this
– hopefully
17 hours ago
My answer is (1/3) arctan (x^2 -1)/sqrt {3} x ..... but the book answer is
– hopefully
17 hours ago
I will include it in the main question above
– hopefully
17 hours ago
But I can see that the final answer in the book contains 2 arctan ...... could you please explain this also for me?
– hopefully
17 hours ago
But I can see that the final answer in the book contains 2 arctan ...... could you please explain this also for me?
– hopefully
17 hours ago
Sure. After substitution, $t=x-frac{1}{x}$, the integral becomes $int frac{1}{1+t^2} , dt$ and integral of this is given by $arctan(t)$
– crskhr
17 hours ago
Sure. After substitution, $t=x-frac{1}{x}$, the integral becomes $int frac{1}{1+t^2} , dt$ and integral of this is given by $arctan(t)$
– crskhr
17 hours ago
No I do not mean this
– hopefully
17 hours ago
No I do not mean this
– hopefully
17 hours ago
My answer is (1/3) arctan (x^2 -1)/sqrt {3} x ..... but the book answer is
– hopefully
17 hours ago
My answer is (1/3) arctan (x^2 -1)/sqrt {3} x ..... but the book answer is
– hopefully
17 hours ago
I will include it in the main question above
– hopefully
17 hours ago
I will include it in the main question above
– hopefully
17 hours ago
|
show 1 more comment
up vote
0
down vote
It turns out that we must compute the integral in the following way:
$$I=intfrac{x^2+1}{x^4+x^2+1}mathrm{d}x$$
$$I=intfrac{x^2+1}{(x^2-x+1)(x^2+x+1)}mathrm{d}x$$
After a fraction decomposition,
$$I=frac12intfrac{mathrm{d}x}{x^2+x+1}+frac12intfrac{mathrm{d}x}{x^2-x+1}$$
Now we focus on
$$I_1=intfrac{mathrm{d}x}{x^2+x+1}$$
Completing the square in the denominator produces
$$I_1=intfrac{mathrm{d}x}{(x+frac12)^2+frac34}$$
Then the substitution $u=frac1{sqrt{3}}(2x+1)$ gives
$$I_1=frac2{sqrt{3}}intfrac{mathrm{d}u}{u^2+1}$$
$$I_1=frac2{sqrt{3}}arctanfrac{2x+1}{sqrt{3}}$$
Similarly,
$$I_2=intfrac{mathrm{d}x}{x^2-x+1}$$
$$I_2=intfrac{mathrm{d}x}{(x-frac12)^2+frac34}$$
And the substitution $u=x-frac12$ carries us to
$$I_2=frac2{sqrt{3}}arctanfrac{2x-1}{sqrt{3}}$$
Plugging in:
$$I=frac1{sqrt{3}}bigg(arctanfrac{2x+1}{sqrt{3}}+arctanfrac{2x-1}{sqrt{3}}bigg)+C$$
Which is confusing, because nowhere did we use the suggested substitution. Furthermore, using the suggested substitution produces
$$I=intfrac{mathrm{d}u}{u^2+3}$$
Which is easily shown to be
$$I=frac1{sqrt{3}}arctanfrac{x^2+1}{xsqrt{3}}$$
But apparently that is incorrect. I am really unsure how this is the case. Try asking your teacher.
answered 4 hours ago
clathratus
1,729219
add a comment |
up vote
0
down vote
It turns out that we must compute the integral in the following way:
$$I=intfrac{x^2+1}{x^4+x^2+1}mathrm{d}x$$
$$I=intfrac{x^2+1}{(x^2-x+1)(x^2+x+1)}mathrm{d}x$$
After a fraction decomposition,
$$I=frac12intfrac{mathrm{d}x}{x^2+x+1}+frac12intfrac{mathrm{d}x}{x^2-x+1}$$
Now we focus on
$$I_1=intfrac{mathrm{d}x}{x^2+x+1}$$
Completing the square in the denominator produces
$$I_1=intfrac{mathrm{d}x}{(x+frac12)^2+frac34}$$
Then the substitution $u=frac1{sqrt{3}}(2x+1)$ gives
$$I_1=frac2{sqrt{3}}intfrac{mathrm{d}u}{u^2+1}$$
$$I_1=frac2{sqrt{3}}arctanfrac{2x+1}{sqrt{3}}$$
Similarly,
$$I_2=intfrac{mathrm{d}x}{x^2-x+1}$$
$$I_2=intfrac{mathrm{d}x}{(x-frac12)^2+frac34}$$
And the substitution $u=x-frac12$ carries us to
$$I_2=frac2{sqrt{3}}arctanfrac{2x-1}{sqrt{3}}$$
Plugging in:
$$I=frac1{sqrt{3}}bigg(arctanfrac{2x+1}{sqrt{3}}+arctanfrac{2x-1}{sqrt{3}}bigg)+C$$
Which is confusing, because nowhere did we use the suggested substitution. Furthermore, using the suggested substitution produces
$$I=intfrac{mathrm{d}u}{u^2+3}$$
Which is easily shown to be
$$I=frac1{sqrt{3}}arctanfrac{x^2+1}{xsqrt{3}}$$
But apparently that is incorrect. I am really unsure how this is the case. Try asking your teacher.
answered 4 hours ago
clathratus
1,729219
add a comment |
up vote
0
down vote
up vote
0
down vote
It turns out that we must compute the integral in the following way:
$$I=intfrac{x^2+1}{x^4+x^2+1}mathrm{d}x$$
$$I=intfrac{x^2+1}{(x^2-x+1)(x^2+x+1)}mathrm{d}x$$
After a fraction decomposition,
$$I=frac12intfrac{mathrm{d}x}{x^2+x+1}+frac12intfrac{mathrm{d}x}{x^2-x+1}$$
Now we focus on
$$I_1=intfrac{mathrm{d}x}{x^2+x+1}$$
Completing the square in the denominator produces
$$I_1=intfrac{mathrm{d}x}{(x+frac12)^2+frac34}$$
Then the substitution $u=frac1{sqrt{3}}(2x+1)$ gives
$$I_1=frac2{sqrt{3}}intfrac{mathrm{d}u}{u^2+1}$$
$$I_1=frac2{sqrt{3}}arctanfrac{2x+1}{sqrt{3}}$$
Similarly,
$$I_2=intfrac{mathrm{d}x}{x^2-x+1}$$
$$I_2=intfrac{mathrm{d}x}{(x-frac12)^2+frac34}$$
And the substitution $u=x-frac12$ carries us to
$$I_2=frac2{sqrt{3}}arctanfrac{2x-1}{sqrt{3}}$$
Plugging in:
$$I=frac1{sqrt{3}}bigg(arctanfrac{2x+1}{sqrt{3}}+arctanfrac{2x-1}{sqrt{3}}bigg)+C$$
Which is confusing, because nowhere did we use the suggested substitution. Furthermore, using the suggested substitution produces
$$I=intfrac{mathrm{d}u}{u^2+3}$$
Which is easily shown to be
$$I=frac1{sqrt{3}}arctanfrac{x^2+1}{xsqrt{3}}$$
But apparently that is incorrect. I am really unsure how this is the case. Try asking your teacher.
answered 4 hours ago
clathratus
1,729219
It turns out that we must compute the integral in the following way:
$$I=intfrac{x^2+1}{x^4+x^2+1}mathrm{d}x$$
$$I=intfrac{x^2+1}{(x^2-x+1)(x^2+x+1)}mathrm{d}x$$
After a fraction decomposition,
$$I=frac12intfrac{mathrm{d}x}{x^2+x+1}+frac12intfrac{mathrm{d}x}{x^2-x+1}$$
Now we focus on
$$I_1=intfrac{mathrm{d}x}{x^2+x+1}$$
Completing the square in the denominator produces
$$I_1=intfrac{mathrm{d}x}{(x+frac12)^2+frac34}$$
Then the substitution $u=frac1{sqrt{3}}(2x+1)$ gives
$$I_1=frac2{sqrt{3}}intfrac{mathrm{d}u}{u^2+1}$$
$$I_1=frac2{sqrt{3}}arctanfrac{2x+1}{sqrt{3}}$$
Similarly,
$$I_2=intfrac{mathrm{d}x}{x^2-x+1}$$
$$I_2=intfrac{mathrm{d}x}{(x-frac12)^2+frac34}$$
And the substitution $u=x-frac12$ carries us to
$$I_2=frac2{sqrt{3}}arctanfrac{2x-1}{sqrt{3}}$$
Plugging in:
$$I=frac1{sqrt{3}}bigg(arctanfrac{2x+1}{sqrt{3}}+arctanfrac{2x-1}{sqrt{3}}bigg)+C$$
Which is confusing, because nowhere did we use the suggested substitution. Furthermore, using the suggested substitution produces
$$I=intfrac{mathrm{d}u}{u^2+3}$$
Which is easily shown to be
$$I=frac1{sqrt{3}}arctanfrac{x^2+1}{xsqrt{3}}$$
But apparently that is incorrect. I am really unsure how this is the case. Try asking your teacher.
answered 4 hours ago
clathratus
1,729219
answered 4 hours ago
clathratus
1,729219
answered 4 hours ago
clathratus
1,729219
answered 4 hours ago
clathratus
1,729219
1,729219
add a comment |
add a comment |
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