Question

Evaluate:
$\int \frac{dx}{x(x^3+8)}$

Answer 1

We have,

$I=\int \frac{dx}{x(x^3+8)}$

$I=\int \frac{x^{2}dx}{x^3(x^3+8)}$

Let $x^3=t\Rightarrow 3x^{2}dx=dt$

$\Rightarrow x^{2}dx=\frac{dt}{3}$

$\therefore$ $I=\int \frac{1}{3t(t+8)}dt$

$=\frac{1}{3}\int \frac{1}{8}\frac{(t+8-t)}{\left(t\right)(t+8)}dt$

$=\frac{1}{24}\int (\frac{1}{t}-\frac{1}{t+8})dt$

$=\frac{1}{24}\log \left(t\right)-log(t+8)+c$

$=\frac{1}{24}\log \left|\frac{t}{t+8}\right|+c$

$=\frac{1}{24}\log \left|\frac{x^3}{x^3+8}\right|+c$

Hence, this is the answer.