Question

Find: $\int \frac{4}{(x-2)(x^2+4)}dx$

Answer 1

$\begin{array}{c}Given,\\ \int \frac{4}{\left(x-2\right)\left(x^2+4\right)}dx\\ Now,\\ 4.\int \frac{1}{\left(x-2\right)\left(x^2+4\right)}dx\\ Applyx=2u\\ 4.\int \frac{1}{4\left(u-2\right)\left(u^2+4\right)}dx\\ \frac{1}{4\left(u-2\right)\left(u^2+4\right)}:\frac{-u-1}{2\left(u^2+1\right)}+\frac{1}{2\left(u-1\right)}\\ =4.\frac{1}{4}\left(\int \frac{-u-1}{2\left(u^2+1\right)}du+\frac{1}{2\left(u-1\right)}du\right)\\ \int \frac{-u-1}{2\left(u^2+1\right)}du=\frac{1}{2}\left(-\frac{1}{2}In\left|u^2+1\right|\right)\\ \int \frac{1}{2\left(u^2+1\right)}du=\frac{1}{2}In\left|u^2+1\right|\\ =4.\frac{1}{4}\left(\frac{1}{2}\left(-\frac{1}{2}In\left|u^2+1\right|\right)-\tan \left(u\right)+\frac{1}{2}\left|u^2+1\right|\right)\\ =\frac{1}{2}\left(-\frac{1}{2}In\left|{\frac{x}{4}}^2+1\right|-\tan \left(\frac{x}{2}\right)+\frac{1}{2}\left|\frac{x}{2}-1\right|\right)\\ =\frac{1}{2}\left(-\frac{1}{2}In\left|{\frac{x}{4}}^2+1\right|-\tan \left(\frac{x}{2}\right)+\frac{1}{2}\left|\frac{x}{2}-1\right|\right)+C\end{array}$