Question
$\int \frac{1}{x\left(x^3-1\right)} d x$
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Answer
$ \text { Let } I =\int \frac{1}{x\left(x^3-1\right)} d x$
$=\int \frac{1}{x \cdot x^3\left(1-\frac{1}{x^3}\right)} d x$
$=\int \frac{1}{x^4\left(1-\frac{1}{x^3}\right)} d x $
Put $1-\frac{1}{x^3}= t$
Differentiating w.r.t.x, we get
$ \frac{3}{x^4} d x= dt$
$\therefore \frac{1}{x^4} d x=\frac{1}{3} dt$
$\therefore \mid=\frac{1}{3} \int \frac{ dt }{ t }$
$=\frac{1}{3} \log | t |+ c$
$=\frac{1}{3} \log \left|1-\frac{1}{x^3}\right|+ c$
$\therefore\left|=\frac{1}{3} \log \right| \frac{x^3-1}{x^3} \mid+ c $
$=\int \frac{1}{x \cdot x^3\left(1-\frac{1}{x^3}\right)} d x$
$=\int \frac{1}{x^4\left(1-\frac{1}{x^3}\right)} d x $
Put $1-\frac{1}{x^3}= t$
Differentiating w.r.t.x, we get
$ \frac{3}{x^4} d x= dt$
$\therefore \frac{1}{x^4} d x=\frac{1}{3} dt$
$\therefore \mid=\frac{1}{3} \int \frac{ dt }{ t }$
$=\frac{1}{3} \log | t |+ c$
$=\frac{1}{3} \log \left|1-\frac{1}{x^3}\right|+ c$
$\therefore\left|=\frac{1}{3} \log \right| \frac{x^3-1}{x^3} \mid+ c $
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