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I'm new here!

The problem: integrate from zero to infinity x over the quantity x cubed plus one dx. I checked on wolfram alpha and the answer is that the indefinite integral is this:

$$\int \frac{x}{1+x^3} dx = \frac{1}{6}\left(\log(x^2-x+1)-2 \log(x+1)+2 \sqrt{3} \arctan((2 x-1)/\sqrt{3})\right)+\text{constant}$$

and the definite integral is this:

$$\int_0^\infty \frac{x}{1+x^3} dx = \frac{2 \pi}{3 \sqrt{3}}\approx 1.2092$$

I am trying to figure out all the steps in between. I see that there are logs, which are equivalent to the ln variant that i am more familiar with, which means it was integrating $1/x$ at some point; I also see an inverse tangent in there.

I started with long division to simplify and I got (which could be wrong because I am very tired right now) $x/(x^3+1) = x^2+ 1/(x^3+1)$ which seems to be a step in the right direction. Wolfram Alpha thinks I definitely did that step wrong. The two equations do not evaluate as equal.

Then I cheated and took the wolfram alpha factorization of $(x^3+1) = (x+1)(x^2-x+1)$...I probably should have known that but didn't offhand. Now it is looking like $x^2$ plus the partial fraction decomposition $1/(x^3+1) = A/(x+1)+(Bx+C)/(x^2-x+1)$. Am I heading in the right direction with this? At this point do I just plug in and crunch?

Winther
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Chad
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  • Welcome to Math.SE. Please edit your question using MathJax for better readability. – Null Oct 17 '14 at 03:34
  • Yes, you're on the right track. The fact that you have three coefficients in your partial fractions decomposition and three (nonconstant) terms in your known antiderivative suggests too that this an efficient way to proceed. – Travis Willse Oct 17 '14 at 03:36

1 Answers1

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You should do something like this:

$$ \displaystyle\frac{x}{x^{3}+1} = \frac{A}{x+1}+\frac{Bx+C}{x^{2}-x+1} $$

$$ \displaystyle \implies Ax^{2}-Ax+A+Bx^{2}+Bx+Cx+C = x $$

$$ \implies (A+B)x^{2}+(-A+B+C)x+(A+C) = x $$

$$ \implies A+B = 0 $$ $$ -A+B+C = 1 $$ $$ A+C = 0 $$

$$ \implies B = C = -A \implies -3A = 1 \implies A = -\frac{1}{3} $$

So $$\int_{0}^{\infty}{\frac{x}{x^{3}+1}}dx = \int_{0}^{\infty}{\frac{-1}{3(x+1)}+\frac{1}{3}\frac{x+1}{x^{2}-x+1}}dx$$

Which is a way easier integral to solve.

Rono
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