Problem source

Show  that if  
$\displaystyle \int\frac1{u \, \f(u)}\; \d u  = \F(u) + c\;$,  
then $\displaystyle \int\frac{m}{x \, \f(x^m)} \;\d x = \F(x^m) + c\;$, 
where $m\ne0$.
 
Find:  
\begin{questionparts} 
\item $\displaystyle\int\frac1{x^n-x} \, \d x\,$; 
\item $\displaystyle\int\frac1 {\sqrt{x^n+x^2}}\, \d x\,$. 
\end{questionparts}

Solution source

\begin{align*}
u = x^m, \d u = m x^{m-1} && \int \frac{m}{x f(x^m)} \d x &= \int \frac{m x^{m-1}}{uf(u)} \d x \\
&&&= \int \frac{1}{u f(u)} \d u \\
&&&= F(u) + c \\
&&&= F(x^m) + c
\end{align*}

\begin{questionparts}
\item \begin{align*}
&& \int \frac{1}{u(u-1)} \d u &=  \int \left ( \frac{1}{u-1}-\frac{1}{u} \right ) \d u \\
&&&= \ln \left ( \frac{u-1}{u} \right) + c \\
&&&= \ln \left ( 1 - \frac{1}{u} \right) + c \\
&& \int \frac{1}{x^n - x} \d x &= \int \frac{1}{x (x^{n-1}-1)} \d x \\
f(u) = u - 1: && &= \frac{1}{n-1} \ln \left ( 1 - \frac{1}{x^{n-1}} \right) + c 
\end{align*}
\item \begin{align*}
v = \sqrt{u+1}, \d v = \tfrac12 (u+1)^{-1/2} \d u && \int \frac{1}{u\sqrt{u+1}} \d u &= \int \frac{1}{(v^2-1)} (u+1)^{-1/2} \d u \\
&&&= \int \frac{2}{v^2-1} \d v \\
&&&=\ln \frac{1-v}{1+v} + c \\
&&&= \ln \left (\frac{1-\sqrt{u+1}}{1+\sqrt{u+1}} \right)+ c \\
f(u) = \sqrt{x+1}:&& \int \frac{1}{\sqrt{x^n + x^2}} \d x &= \int \frac{1}{x\sqrt{x^{n-2}+1}} \d x \\
&&&= \frac{1}{n-2} \ln \left ( \frac{1-\sqrt{x^{n-2}+1}}{1+\sqrt{x^{n-2}+1}} \right)+c
\end{align*}
\end{questionparts}