Integral of The Rational Function

$$ Divide the integral into two parts as the following: $$ $\begin{aligned} \int\frac{x^3+x+2}{x^4+2x^2+1}dx&=\int\frac{x^3+x}{x^4+2x^2+1}dx+\int\frac{2}{x^4+2x^2+1}dx\\ &=\int\frac{x^3+x}{x^4+2x^2+1}dx+\int\frac{2}{(x^2+1)^2}dx. \end{aligned}$ $$ In the RHS part, for the left integral uses substitution $u=x^4+2x^2+1\;\Rightarrow\;du=4(x^3+x)\,dx$ and for the right integral uses substitution $x=\tan\theta\;\Rightarrow\;dx=\sec^2\theta\;d\theta$ . Therefore $$ $\begin{aligned} \int\frac{x^3+x+2}{x^4+2x^2+1}dx&=\int\frac{1}{4u}du+\int\frac{2\sec^2\theta}{(\tan^2\theta+1)^2}d\theta\\ &=\frac{1}{4}\ln\,|u|+\text{C}+2\int\frac{\sec^2\theta}{\sec^4\theta}d\theta\\ &=\frac{1}{4}\ln\,\left|x^4+2x^2+1\right|+2\int\cos^2\theta\;d\theta+\text{C}\\ &=\frac{1}{4}\ln\,\left|x^4+2x^2+1\right|+2\int\frac{1+\cos2\theta}{2}d\theta+\text{C}\\ &=\frac{1}{4}\ln\,\left|x^4+2x^2+1\right|+\int(1+\cos2\theta)\;d\theta+\text{C}\\ \end{aligned}$ $$ The rest should be easy to be solved and I leave to you as an exercise. #LOL $$

$\text{\# }\mathbb{Q.E.D.}\text{ \#}$

Break as $\displaystyle \int_{0}^{1} \frac{x(x^2+1)}{(x^2+1)^2} dx + \int_{0}^{1} \frac{1-x^2+1+x^2}{(1+x^2)^2} dx$

= $\displaystyle \int_{0}^{1} \frac{x dx}{x^2+1} + \int_{0}^{1} \frac{1}{1+x^2} dx - \int_{0}^{1} \frac{1 -1/x^2}{(x+1/x)^2} dx$

= $\displaystyle \frac{\ln 2}{2} + \frac{\pi}{4} + \int_{2}^{\infty} \frac{dt}{t^2}$ , where $t = x+1/x$

Hence, $I = \frac{\ln 2}{2} + \frac{\pi}{4} + \frac{1}{2}$

You have to replace $\frac{\ln 2}{2}$ by $\frac{\ln 4}{4}$ so as to make $t$ and $u$ perfect squares.

very nice solution Tunk - Fey