Question

integrate ((x+1)/(x^2+1))^2

Answer 1

\[ \begin{align*}
\int \left(\frac{x+1}{x^2+1} \right )^2 dx &= \int \frac{(x+1)^2}{x^2 ( x + 1/x)}dx\\
&= \int \frac{1/x^2 + 2/x + 1}{(x+1/x)^2 }dx \\
&= \int \frac{-1+1/x^2 }{(x+1/x)^2} dx + 2\int \frac{1/x + 1}{(x+1/x)^2}dx\\
&= \frac{-x}{x^2 + 1 } + \int ... \\

\end{align*} \]

Answer 2

\[
\begin{align*}
\int \frac{1/x + 1}{(x+1/x)^2}dx &= \int \frac{\tan u + 1}{ (1 + \tan^2 u )^2}\sec^2u du\\
&= \int \frac{\tan u + 1}{\sec^2 u } du \\
&= \int \sin u \cos u du + \int \cos^2u du \\
&= \frac 1 2 \int \sin 2u du + \frac 1 2 \int (\cos 2u + 1 )du
\end{align*} \]

Answer 3

seems like the usual trick would have worked initially.

Answer 4

starting off with a trig substitution would work here as well
x = tan(u)
dx = sec^2 (u) du
\[\rightarrow \int\limits \frac{(\tan u +1)^{2}}{\sec^{4} u} \sec^{2} u du\]
\[= \int\limits \frac{\sec^{2} u +2\tan u}{\sec^{2} u} du\]
\[= \int\limits 1 + 2\sin u \cos u du\]
\[= \int\limits 1 + \sin (2u) du\]
\[ = u -\frac{1}{2} \cos(2u) +C\]
\[=u - \frac{1}{2}(2\cos^{2} u -1) + C\]
\[= u - \cos^{2} u + C\]
\[= \tan^{-1} x - \frac{1}{x^{2} +1} +C\]