Integrate the following functions w.r.t. x: 1 x (x^3-1 ) - Vidyadip

Question

Integrate the following functions w.r.t. x:
$\frac{1}{x\left(x^3-1\right)}$

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Answer

Let $I=\int \frac{1}{x\left(x^3-1\right)} d x$
$=\int \frac{x^{-4}}{x^{-4} x\left(x^3-1\right)} d x$
$=\int \frac{x^{-4}}{1-x^{-3}} d x$
$=\frac{1}{3} \int \frac{3 x^{-4}}{1-x^{-3}} d x$
$=\frac{1}{3} \int \frac{\frac{d}{d x}\left(1-x^{-3}\right)}{1-x^{-3}} d x$
$=\frac{1}{3} \log \left|1-x^{-3}\right|+c$
$\cdots\left[\because \int \frac{f^{\prime}(x)}{f(x)} d x=\log |f(x)|+c\right]$
$=\frac{1}{3} \log \left|1-\frac{1}{x^3}\right|+c$
$=\frac{1}{3} \log \left|\frac{x^3-1}{x^3}\right|+c .$
Alternative Method:
$\text { Let } I=\int \frac{1}{x\left(x^3-1\right)} d x$
$=\int \frac{x^2}{x^3\left(x^3-1\right)} d x$
Put $x^3=t \quad \therefore 3 x^2 d x=d t$
$\therefore x^2 d x=\frac{d t}{3}$
$\therefore I =\int \frac{1}{t(t-1)} \cdot \frac{d t}{3}=\frac{1}{3} \int \frac{1}{t(t-1)} d t$
$ =\frac{1}{3} \int \frac{t-(t-1)}{t(t-1)} d t=\frac{1}{3} \int\left(\frac{1}{t-1}-\frac{1}{t}\right) d t$
$ =\frac{1}{3}\left[\int \frac{1}{t-1} d t-\int \frac{1}{t} d t\right]$
$=\frac{1}{3}[\log |t-1|-\log |t|]+c$
$=\frac{1}{3} \log \left|\frac{t-1}{t}\right|+c$
$=\frac{1}{3} \log \left|\frac{x^3-1}{x^3}\right|+c .$

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