How to evaluate the following indefinite integral? $\int\frac{1}{x(x^2-1)}dx.$ I need the step by step solution of this integral
please help me! 
I can't solve it! 
$$\int\frac{1}{x(x^2-1)}dx.$$

We use partial fraction decomposition:
$$\int\frac{1}{x(x^2-1)}dx = \int \frac 1{x(x-1)(x+1)}\,dx = \int \left(\frac A{x} + \frac{B}{x - 1} + \frac C{x+1}\right)\,dx$$
Solving for $A, B, C$:
$$A(x-1)(x+1) + Bx(x+1) + Cx(x-1) = 1$$
When $x = 1 \implies 2B = 1 \implies B = \frac 12$
$x = -1 \implies 2C = 1 \iff C = \frac 12$ 
$x = 0 \implies -A = 1 \iff A = -1$.
That gives us: $$\int \left(\frac {-1}{x} + \frac{1}{2(x - 1)} + \frac 1{2(x+1)}\right)\,dx$$
Now use the fact that $\int \frac{f'(x)}{f(x)}\,dx = \ln|f(x)| + C$.
