Question

Find $\int \frac{2x}{\left(x^2+1\right){\left(x^2+2\right)}^2}dx$

Answer 1

$\int \frac{2x}{\left(x^2+1\right){\left(x^2+2\right)}^2}dx$

$$\begin{gathered} \mathrm{Let}{x}^2=y \\ \Rightarrow 2x\mathrm{d}x=\mathrm{d}y \\ \Rightarrow \mathrm{d}x=\frac{\mathrm{d}y}{2x} \end{gathered}$$

$$\begin{gathered} \int \frac{2x}{\left(x^2+1\right){\left(x^2+2\right)}^2}dx \\ =\int \frac{dy}{\left(y+1\right){\left(y+2\right)}^2} \\ \mathrm{Let}\frac{1}{\left(y+1\right){\left(y+2\right)}^2}=\frac{A}{y+1}+\frac{B}{y+2}+\frac{C}{{\left(y+2\right)}^2}.....\left(1\right) \\ \Rightarrow 1=A{\left(y+2\right)}^2+B\left(y+1\right)\left(y+2\right)+C\left(y+1\right).....\left(2\right) \\ \mathrm{Putting}y=-2\mathrm{in}\left(2\right) \\ 1=C\left(-2+1\right) \\ \Rightarrow C=-1 \end{gathered}$$

$$\begin{gathered} \mathrm{Putting}y=-1\mathrm{in}\left(2\right) \\ 1=A{\left(-1+2\right)}^2 \\ \Rightarrow 1=A\left(1\right) \\ \Rightarrow A=1 \end{gathered}$$

$$\begin{gathered} \mathrm{Putting}y=0\mathrm{in}\left(2\right) \\ 1=4A+B\left(2\right)+C \\ \Rightarrow 1=4+2B-1 \\ \Rightarrow 1=3+2B \\ \Rightarrow -2=2B \\ \Rightarrow B=-1 \end{gathered}$$

Substituting the values of A, B and C in (1)

$$\begin{gathered} \frac{1}{\left(y+1\right){\left(y+2\right)}^2}=\frac{1}{y+1}-\frac{1}{y+2}-\frac{1}{{\left(y+2\right)}^2} \\ \Rightarrow \int \frac{dy}{\left(y+1\right){\left(y+2\right)}^2}=\int \frac{dy}{y+1}-\int \frac{dy}{y+2}-\int \frac{dy}{{\left(y+2\right)}^2} \\ =\log \left|y+1\right|-\log \left|y+2\right|+\frac{1}{y+2}+C \end{gathered}$$
Hence, $\int \frac{2x}{\left(x^2+1\right){\left(x^2+2\right)}^2}dx$ = $=\log \left|x^2+1\right|-\log \left|x^2+2\right|+\frac{1}{x^2+2}+C$