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! 
$$\int\frac{1}{x(x^2-1)}dx.$$
 A: 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 $\int \frac{f'(x)}{f(x)}\,dx = \ln|f(x)| + C$.
A: Hint:  Use partial fraction decomposition to prove that: $$\dfrac1{x(x^2-1)}=-\dfrac{1}{x}+\dfrac{1}{2(x+1)}+\dfrac1{2(x-1)}.$$
The rest is straightforward. 
A: $$I=\int\frac{dx}{x(x^2-1)}=\int\frac{x\ dx}{x^2(x^2-1)}$$
Setting $x^2=y,2x\ dx=dy$
$$2I=\int\frac{dy}{y(y-1)}=\int\frac{\{y-(y-1)\}dy}{y(y-1)}=\int\frac{dy}{y-1}-\int\frac{dy}y$$
$$=\ln|y-1|-\ln |y|+K$$
$$=\ln|x^2-1|-\ln |x^2|+K$$
$$2I=\ln|x^2-1|-2\ln |x|+K$$
A: Partial fractions always works.  However, it looks like the fastest way might be to multiply top and bottom by $x^{-3}$.
$$\int\frac{dx}{x(x^2-1)}=\int\frac{x^{-3}dx}{xx^{-1}[x^{-2}(x^2-1)]}=\frac12\int\frac{2x^{-3}dx}{1-x^{-2}}=\frac12\ln|1-x^{-2}|+C$$
A: 1/{x*(x^2-1)}
=x/{x^2*(x^2-1)}
If we substitute:
 x^2=z
By differentiating both sides
2x dx = dz
x dx= dz/2
Now if we solve the integral
(1/2)log{(x^2-1)/x^2}+C
A: $\newcommand{\+}{^{\dagger}}
 \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
 \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
 \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,}
 \newcommand{\dd}{{\rm d}}
 \newcommand{\down}{\downarrow}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}
 \newcommand{\fermi}{\,{\rm f}}
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{{\rm i}}
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\isdiv}{\,\left.\right\vert\,}
 \newcommand{\ket}[1]{\left\vert #1\right\rangle}
 \newcommand{\ol}[1]{\overline{#1}}
 \newcommand{\pars}[1]{\left(\, #1 \,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,}
 \newcommand{\sech}{\,{\rm sech}}
 \newcommand{\sgn}{\,{\rm sgn}}
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}
 \newcommand{\wt}[1]{\widetilde{#1}}$
\begin{align}&\overbrace{\color{#66f}{\large\int{\dd x \over x\pars{x^{2} - 1}}}}
^{\ds{\mbox{Set}\ x \equiv \sec\pars{\theta}}}\ =\
\int{\sec\pars{\theta}\tan\pars{\theta}\,\dd\theta \over \sec\pars{\theta}\tan^{2}\pars{\theta}}
=\int{\dd\theta \over \tan\pars{\theta}}
=\int{\cos\pars{\theta}\,\dd\theta \over \sin\pars{\theta}}
\\[3mm]&=\ln\pars{\sin\pars{\theta}}
=\ln\pars{\tan\pars{\theta} \over \sec\pars{\theta}}
=\ln\pars{\root{\sec^{2}\pars{\theta} - 1} \over \sec\pars{\theta}}
\\[3mm]&=\color{#66f}{\large\ln\pars{\root{x^{2} - 1} \over x} + \mbox{a constant.}}
\end{align}
