Question

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

Answer 1

Let $t=x^2+1\Rightarrow dt=2xdx$

$\Rightarrow t+1=x^2+2$

$\int \frac{2xdx}{(x^2+1){(x^2+2)}^2}$

$\int \frac{dt}{t{(t+1)}^2}$

Consider $\frac{1}{t{(t+1)}^2}$ Let us solve this by the method of partial fractions,

$\frac{1}{t{(t+1)}^2}=\frac{A}{t}+\frac{B}{(t+1)}+\frac{C}{{(t+1)}^2}$

$\Rightarrow 1=A{(t+1)}^2+Bt(t+1)+Ct$ for $A,B,C\in R$

Put $t=0\Rightarrow 1=A$

$\therefore A=1$

Put $t=-1\Rightarrow 1=A\left(0\right)+B\times 0+c\times -1$

$\therefore C=-1$

Put $t=1\Rightarrow 2A+2B+C=1$

$\Rightarrow 2+2B-1=1$ where $A=1,C=-1$

$\Rightarrow 2B=1-1=0$

$\therefore \int \frac{dt}{t{(t+1)}^2}=\int (\frac{A}{t}+\frac{B}{(t+1)}+\frac{C}{{(t+1)}^2})dt$

$\Rightarrow \int \frac{dt}{t{(t+1)}^2}=\int (\frac{1}{t}+\frac{0}{(t+1)}+\frac{-1}{{(t+1)}^2})dt$

$=\int \frac{dt}{t}-\int \frac{dt}{{(t+1)}^2}$

$=\ln \left|t\right|-\frac{{(t+1)}^{-2+1}}{-2+1}+c$ using $\int x^{n}dx=\frac{x^{n+1}}{n+1}$ and $\int \frac{dx}{x}=\ln \left|x\right|$

$=\ln \left|t\right|-\frac{1}{t+1}+c$ where $c$ is the constant of integration.

$=\ln |x^2+1|-\frac{1}{x^2+1+1}+c$

$=\ln |x^2+1|-\frac{1}{x^2+2}+c$