Question

$$\int \frac{5x^2-1}{x(x-1)(x+1)}dx$$

Answer 1

We have,
$$I=\int \frac{(5x^2-1)dx}{x(x-1)(x+1)}$$

$$\text{Let }\frac{5x^2-1}{x(x-1)(x+1)}=\frac{A}{x}+\frac{B}{x-1}+\frac{C}{x+1}$$

$$\Rightarrow \frac{5x^2-1}{x(x-1)(x+1)}=\frac{A(x^2-1)+Bx\cdot (x+1)+C\cdot x\cdot (x-1)}{x(x-1)(x+1)}$$

$$\Rightarrow 5x^2-1=A(x^2-1)+B\cdot x(x+1)+C\cdot x\cdot (x-1)$$

Putting x = 1

$$\Rightarrow 5-1=A\times 0+B\left(1\right)(1+1)+C\times 0$$

$$\Rightarrow 4=B\left(2\right)$$

$$\Rightarrow B=2$$

Putting x = 0

$$\Rightarrow 5\times 0-1=A(0-1)+B\times 0+C\times 0$$

$$\Rightarrow -1=A(-1)$$

$$\Rightarrow A=1$$

Putting x + 1 = 0

$$x=-1$$

$$5-1=A\times 0+B\times 0+C(-1)(-2)$$

$$\Rightarrow C=2$$

$$\therefore I=\int \frac{dx}{x}+2\int \frac{dx}{x-1}+2\int \frac{dx}{x+1}$$

$$=\log \left|x\right|+2\log |x-1|+2\log |x+1|+C$$

$$=\log \left|x\right|+2\{\log |x^2-1|\}+C$$

$$=\log \left|x{(x^2-1)}^2\right|+C$$