Integrate: $\int \:x\left(\frac{1-x^2}{1+x^2}\right)^2dx$.

Question

Integrate: $$\int \:x\left(\frac{1-x^2}{1+x^2}\right)^2dx$$

My attempt:

$$\text{Let} \ u = x, v'=\left(\frac{1-x^2}{1+x^2}\right)^2\\$$ \begin{align} \int \:x\left(\frac{1-x^2}{1+x^2}\right)^2dx & = x\left(x-2\arctan \left(x\right)+\frac{2x}{1+x^2}\right)-\int \frac{3x+x^3-2\arctan \left(x\right)-2x^2\arctan \left(x\right)}{1+x^2}dx\\ & = x\left(x-2\arctan \left(x\right)+\frac{2x}{1+x^2}\right)-\left(\frac{1}{2}x^2-2x\arctan \left(x\right)+\ln \left|x^2+1\right|-2\ln \left|\frac{1}{\sqrt{x^2+1}}\right|+\frac{1}{2}\right)\\ & = \frac{1}{2}x^2+\frac{2x^2}{x^2+1}-\ln \left|x^2+1\right|+2\ln \left|\frac{1}{\sqrt{x^2+1}}\right|-\frac{1}{2}+C,C \in \mathbb{R} \end{align}

I left out the small details, otherwise this post would be quite long. I tried to do this with $u$-substitution but I'm not sure how I can do that here.

Answer 1

Let $x^2=t\implies xdx=\frac{dt}{2}$ $$\int x\left(\frac{1-x^2}{1+x^2}\right)^2dx=\int \left(\frac{1-t}{1+t}\right)^2\frac{dt}{2}$$ $$=\frac12\int \left(\frac{2}{1+t}-1\right)^2 dt$$ $$=\frac12\int \left(1+\frac{4}{(1+t)^2}-\frac{4}{1+t}\right) dt$$ $$=\frac12 \left(t-\frac{4}{1+t}-4\ln|1+t|\right)+C$$ $$=\frac{x^2}{2}-\frac{2}{1+x^2}-2\ln(1+x^2)+C$$

Answer 2

Hint:

$$1+x^2=t\implies2x\ dx=dt\text{ and }1-x^2=1-(t-1)=?$$

$$\int x\left(\dfrac{1-x^2}{1+x^2}\right)^2dx=\dfrac12\int\left(\dfrac{2-t}t\right)^2dt$$

Answer 3

Another easier way to integrate as follows $$\int x\left(\frac{1-x^2}{1+x^2}\right)^2dx$$$$=\int \left(\frac{2}{1+x^2}-1\right)^2xdx$$ $$=\frac12\int \left(\frac{4}{(1+x^2)^2}+1-\frac{4}{1+x^2}\right)d(1+x^2)$$ $$=\frac12\left(-\frac{4}{1+x^2}+1+x^2-4\ln(1+x^2)\right)+c$$ $$=\bbox[15px,#ffd,border:1px solid green]{\frac{x^2}{2}-\frac{2}{1+x^2}-2\ln(1+x^2)+C}$$

Answer 4

Let $x=\tan(t)$: $$\int x\left(\frac{1-x^2}{1+x^2}\right)^2\,dx$$ $$\Rightarrow \int \tan(t)\left(\frac{1-\tan^2(t)}{\sec^2(t)}\right)^2\sec^2(t)\,dt$$ $$ = \int \sin^2(t) \tan^3(t) - 2 \sin^2(t)\tan(t) + \sin(t) \cos(t)\,dt $$ $$ = \frac{1}{2}\sec^2(t)+2\sin^2(t) -2 \log(\sec^2(t)) +C $$ $$ \Rightarrow \frac{1}{2}x^2 +2 \frac{x^2}{1 + x^2} - 2 \log(1 + x^2)+C $$

Answer 5

$$\int x({1-x^2\over 1+x^2})^2 dx$$

substitute $1+x^2=u$,

$x^2=u-1$,

$2xdx=du$, $$\int ({1-(u-1)\over u})^2 {du\over 2}$$ $$=\int ({2-u\over u})^2 {du\over 2}$$ use:$(x-y)^2=x^2+y^2-2xy$ $$={1\over 2}\int ({2\over u}-1)^2 du$$$$={1\over 2}\int {4\over u^2}+1-{4\over u} du$$ $$={1\over 2}({-4\over u}+u-4\ln|u|)+c$$ $$={-2\over u}+{u\over 2}-2\ln|u|+c$$ $$={x^2\over 2}-{2\over 1+x^2}-2\ln|1+x^2|+c$$