Question

$\int \frac{18}{\left(x+2\right)\left(x^2+4\right)}dx$

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ I=\int \frac{18}{\left(x+2\right)\left(x^2+4\right)}dx \\ \mathrm{Let}\frac{18}{\left(x+2\right)\left(x^2+4\right)}=\frac{A}{x+2}+\frac{Bx+C}{x^2+4} \\ \Rightarrow \frac{18}{\left(x+2\right)\left(x^2+4\right)}=\frac{A\left(x^2+4\right)+\left(Bx+C\right)\left(x+2\right)}{\left(x+2\right)\left(x^2+4\right)} \\ \Rightarrow 18=A{x}^2+4A+B{x}^2+2Bx+Cx+2C \\ \Rightarrow 18=\left(A+B\right)x^2+x\left(2B+C\right)+4A+2C \\ \mathrm{Equating}\mathrm{coefficients}\mathrm{of}\mathrm{like}\mathrm{terms} \\ A+B=0.....\left(1\right) \\ 2B+C=0.....\left(2\right) \\ 4A+2C=18.....\left(3\right) \\ \mathrm{Solving}\left(1\right),\left(2\right)\mathrm{and}\left(3\right),\mathrm{we}\mathrm{get} \\ \mathrm{A}=\frac{9}{4} \\ \mathrm{B}=-\frac{9}{4} \\ \mathrm{C}=\frac{9}{2} \\ \therefore \frac{18}{\left(x+2\right)\left(x^2+4\right)}=\frac{9}{4\left(x+2\right)}+\frac{-{\displaystyle \frac{9}{4}}x+{\displaystyle \frac{9}{2}}}{x^2+4} \\ \Rightarrow \frac{18}{\left(x+2\right)\left(x^2+4\right)}=\frac{9}{4\left(x+2\right)}-\frac{9}{4}\left(\frac{x}{x^2+4}\right)+\frac{9}{2\left(x^2+4\right)} \\ \Rightarrow \int \frac{18dx}{\left(x+2\right)\left(x^2+4\right)}=\frac{9}{4}\int \frac{dx}{x+2}-\frac{9}{4}\int \frac{xdx}{x^2+4}+\frac{9}{2}\int \frac{dx}{x^2+2^2} \\ \mathrm{let}{x}^2+4=t \\ \Rightarrow 2xdx=dt \\ \Rightarrow xdx=\frac{dt}{2} \\ \therefore I=\frac{9}{4}\int \frac{dx}{x+2}-\frac{9}{8}\int \frac{dt}{t}+\frac{9}{2}\int \frac{dx}{x^2+2^2} \\ =\frac{9}{4}\log \left|x+2\right|-\frac{9}{8}\log \left|t\right|+\frac{9}{2}\times \frac{1}{2}{\tan }^{-1}\left(\frac{x}{2}\right)+C\text{'} \\ =\frac{9}{4}\log \left|x+2\right|-\frac{9}{8}\log \left|x^2+4\right|+\frac{9}{4}{\tan }^{-1}\left(\frac{x}{2}\right)+C\text{'} \end{gathered}$$