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Calculate using integration by parts $$\int \dfrac{x^2}{(x^2+1)^2}$$

I'm looking through some working for this question and it gives

$u=x/2, u'=1/2$

$v'=\dfrac{2x}{(x^2+1)^2}, v=\dfrac{-1}{x^2+1}$

I'm confused as to why these values for u and v are used and would appreciate it if someone could explain why

Jim
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  • Having $u=\dfrac{x}{2}$ and $v'=\dfrac{2x}{(x^2+1)^2}$ reduces the power of the integrand. The solution given chooses to multiply the integral by $\dfrac{2}{2}$ because it prevents a factor of $\dfrac{1}{2}$ from showing up later, I think. In any case, using either $u=\dfrac{x}{2}$ or $u=x$ will lead to the same solution. – user170231 Oct 29 '15 at 14:10
  • Isn't u = x points as you'd get the same integral with only a change in variable name? – fleablood Oct 29 '15 at 14:14
  • @fleablood, $u=x$ gives the same integral if you are doing a regular substitution, but this is an integration by parts. – vadim123 Oct 29 '15 at 14:21
  • They are chosen because $v'$ can be easily integrated by substitution $w = x^2 + 1$. – Matt Dickau Oct 29 '15 at 14:24
  • These values are used because you want the first function which is differentiable and the second function which is integrable – Shailesh Oct 29 '15 at 14:36

1 Answers1

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Notice, $$\int \frac{x^2}{(x^2+1)^2}\ dx$$$$=\frac{1}{2}\int \underbrace{x}_{I}\cdot \underbrace{\frac{2x}{(x^2+1)^2}}_{II}\ dx$$ Now, using integration by parts, $$=\frac{1}{2}\left(x\int \frac{2x}{(x^2+1)^2}\ dx-\int \left(\frac{d}{dx}(x)\cdot \int \frac{2x}{(x^2+1)^2}\ dx\right) dx\right)$$ $$=\frac{1}{2}\left(x\int \frac{d(x^2+1)}{(x^2+1)^2}-\int \left(\frac{d}{dx}(x)\cdot \int \frac{d(x^2+1)}{(x^2+1)^2}\right)\ dx\right)$$ $$=\frac{1}{2}\left(x\left(-\frac{1}{x^2+1}\right)-\int 1\cdot \left(-\frac{1}{x^2+1}\right)\ dx\right)$$

$$=\frac{1}{2}\left(-\frac{x}{x^2+1}\right)-\frac{1}{2}\int \left(-\frac{1}{x^2+1} \right) dx$$ $$=-\frac{x}{2(x^2+1)}+\frac{1}{2}\int \frac{1}{x^2+1}\ dx$$ $$=-\frac{x}{2(x^2+1)}+\frac{1}{2}\tan^{-1}(x)+C$$

Harish Chandra Rajpoot
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