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

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

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Surely you tried partial fractions? –  MPW Jun 19 at 11:22

6 Answers 6

up vote 9 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 $\int \frac{f'(x)}{f(x)}\,dx = \ln|f(x)| + C$.

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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.

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

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

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

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

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

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@MelikaE, How about this? –  lab bhattacharjee Jun 19 at 11:46

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

$$\int\frac{dx}{x(x^2-1)}=\int\frac{x^{-3}dx}{xx^{-1}[x^{-2}(x^2-1)]}=\frac12\int\frac{2x^{-3}dx}{1-x^{-2}}=\frac12\ln|1-x^{-2}|+C$$

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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

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

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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 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 at 21:56

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