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I'm trying to calculate $\int \frac{x^2 }{(x^2+1)^2} dx$ by using the formula: $$ \int udv = uv -\int vdu $$

I supposed that $u=x$ s.t $du=dx$, and also that $dv=\frac{x}{(x^2+1)^2}dx$, but I couldn't calculate the last integral. what is the tick here?

the answer must be: $ \frac{1}{2}arctan\ x - \frac{x}{2(1+x^2)} $ + C

Firas Abd El Gani
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3 Answers3

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First, you used parts: with $u = x$ and $dv = \frac{x}{(x^2+1)^2}dx$, you got $$\int\frac{x^2}{(x^2+1)^2}dx = xv - \int v\,dx.\tag{1}$$ From here, I suggested the substitution $t = x^2$ in order to integrate $dv$ and get $v$. Doing this, we see that $dt = 2x\,dx$, so $x\,dx = \frac{1}{2}dt$. Thus $$v = \int dv = \int \frac{x}{(x^2+1)^2}dx = \frac{1}{2} \int \frac{dt}{(t + 1)^2}.$$ This latter integral is easy: if you can't guess it, let $s = t + 1$. In any case, we get $$v = -\frac{1}{2(t+1)} = -\frac{1}{2(x^2+1)}.$$ Going back to $(1)$, $$\int\frac{x^2}{(x^2+1)^2}dx = -\frac{x}{2(x^2+1)} + \frac{1}{2} \int \frac{dx}{x^2+1}$$ and you should recognize the final integral as $\arctan{x}$. Putting it all together, the answer is $$\frac{1}{2}\arctan{x} - \frac{x}{2(x^2+1)} + C.$$

Unit
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HINT:

Use trigonometric substitution $x=\tan\theta$

$\int\dfrac{x^2}{(x^2+1)^2}dx=\cdots=\dfrac12\int(1-\cos2\theta)d\theta$

$\sin2\theta=\dfrac{2\tan\theta}{1+\tan^2\theta}$

lab bhattacharjee
  • 274,582
  • hmm, I'm not sure that I learned this trigo equation, but It lookds like a nice idea. thats for the effort. i will be glad if u can solve it in other way, or at least, according to this formula. – Firas Abd El Gani Mar 20 '15 at 18:07
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    @FirasAliAbdelGhani, See http://www.sosmath.com/calculus/integration/trigsub/trigsub.html – lab bhattacharjee Mar 20 '15 at 18:14
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$$\int\dfrac{x^2}{(x^2+1)^2}dx$$

$$=x\int\dfrac x{(x^2+1)^2}dx-\int\left[\frac{dx}{dx}\cdot\int\dfrac x{(x^2+1)^2}dx\right]dx$$

$$=x\cdot\frac{-1}{1+x^2}+\int\dfrac{dx}{1+x^2}=\cdots$$

lab bhattacharjee
  • 274,582