Problem source

Calculate the following integrals

\begin{questionparts}
\item ${\displaystyle \int\frac{x}{(x-1)(x^{2}-1)}\,\mathrm{d}x}$; 
\item ${\displaystyle \int\frac{1}{3\cos x+4\sin x}\,\mathrm{d}x}$; 
\item ${\displaystyle \int\frac{1}{\sinh x}\,\mathrm{d}x}.$ 
\end{questionparts}

Solution source

\begin{questionparts}
\item \begin{align*}
\int\frac{x}{(x-1)(x^{2}-1)}\,\mathrm{d}x &= \int \frac{x}{(x-1)^2 (x+1)} \d x \\
&= \int \frac{1}{2(x-1)^2} + \frac{1}{4(x-1)} - \frac{1}{4(x+1)} \d x \\
&= -\frac12 (x-1)^{-1} + \frac14 \ln(x-1) - \frac14 \ln (x+1) + C
\end{align*}
\item \begin{align*}
\int \frac{1}{3 \cos x + 4 \sin x } \d x &= \int \frac{1}{5 \cos (x - \cos^{-1}(3/5))} \d x \\
&= \frac15 \int \sec (x - \cos^{-1}(3/5)) \d x\\
&= \frac15 \left (\ln | \sec (x - \cos^{-1}(3/5))  + \tan (x - \cos^{-1}(3/5)) | \right) + C
\end{align*}
\item \begin{align*}
\int \frac{1}{\sinh x} \d x &= \int \frac{2}{e^x - e^{-x}} \\
&= \int \frac{2e^x}{e^{2x}-1} \d x \\
&=\int \frac{e^x}{e^x-1} - \frac{e^x}{e^x+1} \d x \\
&= \ln (e^x - 1) + \ln (e^x+1) + C
\end{align*}
\end{questionparts}