Prove
Question:

$\frac{x^{2}}{\left(2+3 x^{3}\right)^{3}}$

Solution:

Let $2+3 x^{3}=t$

$\therefore 9 x^{2} d x=d t$
$\Rightarrow \int \frac{x^{2}}{\left(2+3 x^{3}\right)^{3}} d x=\frac{1}{9} \int \frac{d t}{(t)^{3}}$

$=\frac{1}{9}\left[\frac{t^{-2}}{-2}\right]+\mathrm{C}$

$=\frac{-1}{18}\left(\frac{1}{t^{2}}\right)+\mathrm{C}$

$=\frac{-1}{18\left(2+3 x^{3}\right)^{2}}+\mathrm{C}$

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