Question

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

Answer 1

We have,

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

$$\text{Let }\frac{x}{(x+1)(x^2+1)}=\frac{A}{x+1}+\frac{Bx+C}{x^2+1}$$

$$\Rightarrow \frac{x}{(x+1)(x^2+1)}=\frac{A(x^2+1)+(Bx+C)(x+1)}{(x+1)(x^2+1)}$$

$$\Rightarrow x=A(x^2+1)+B{x}^2+Bx+Cx+C$$

$$\Rightarrow x=(A+B)x^2+(B+C)x+(A+C)$$

$$\text{Equating coefficients of like terms}$$

$$A+B=0.....\left(1\right)$$

$$B+C=1.....\left(2\right)$$

$$A+C=0.....\left(3\right)$$

$$\text{Solving (1), (2) and (3), we get}$$

$$A=-\frac{1}{2}$$

$$B=\frac{1}{2}$$

$$C=\frac{1}{2}$$

$$\therefore \frac{x}{(x+1)(x^2+1)}=-\frac{1}{2(x+1)}+\frac{\frac{x}{2}+\frac{1}{2}}{x^2+1}$$

$$\Rightarrow \int \frac{xdx}{(x+1)(x^2+1)}=-\frac{1}{2}\int \frac{dx}{x+1}+\frac{1}{2}\int \frac{xdx}{x^2+1}+\frac{1}{2}\int \frac{dx}{x^2+1}$$

$$\text{Let }{x}^2+1=t$$

$$\Rightarrow 2xdx=dt$$

$$\Rightarrow xdx=\frac{dt}{2}$$

$$\therefore I=-\frac{1}{2}\int \frac{dx}{x+1}+\frac{1}{4}\int \frac{dt}{t}+\frac{1}{2}\int \frac{dx}{x^2+1^2}$$

$$=-\frac{1}{2}\log |x+1|+\frac{1}{4}\log \left|t\right|+\frac{1}{2}{\tan }^{-1}x+C^{\prime }$$

$$=-\frac{1}{2}\log |x+1|+\frac{1}{4}\log |x^2+1|+\frac{1}{2}{\tan }^{-1}x+C^{\prime }$$