Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I need the step by step solution of this integral please help me! I can't solve it!


share|cite|improve this question
Surely you tried partial fractions? – MPW Jun 19 '14 at 11:22
up vote 7 down vote accepted

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 $\int \frac{f'(x)}{f(x)}\,dx = \ln|f(x)| + C$.

share|cite|improve this answer

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.

share|cite|improve this answer

$$I=\int\frac{dx}{x(x^2-1)}=\int\frac{x\ dx}{x^2(x^2-1)}$$

Setting $x^2=y,2x\ dx=dy$

$$2I=\int\frac{dy}{y(y-1)}=\int\frac{\{y-(y-1)\}dy}{y(y-1)}=\int\frac{dy}{y-1}-\int\frac{dy}y$$ $$=\ln|y-1|-\ln |y|+K$$

$$=\ln|x^2-1|-\ln |x^2|+K$$

$$2I=\ln|x^2-1|-2\ln |x|+K$$

share|cite|improve this answer
@MelikaE, How about this? – lab bhattacharjee Jun 19 '14 at 11:46

Partial fractions always works. However, it looks like the fastest way might be to multiply top and bottom by $x^{-3}$.


share|cite|improve this answer

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

share|cite|improve this answer

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ \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}

share|cite|improve this answer
Sir, with all due respect, your answer is right but I'm not so sure about the substitution. By letting $x= \sec \theta$ aren't you confining the set of values of $x$ can take, even though there's no such constraint on $x$? (since $\sec$ can't take values between -1 and 1) – Mathguy Jul 2 '14 at 21:53
$\large x$ can be 'even' a complex number. Additional details are necessary whenever the integral becomes a definite one such as sign,etc.. – Felix Marin Jul 2 '14 at 21:56

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.