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Q. $\int\left(27e^{9x}+e^{12x}\right)^{1/3}dx$ is equal to

KEAMKEAM 2013Integrals

Solution:

$ I=\int\left(27 e^{9 x}+e^{12 x}\right)^{1 / 3} d x $
$=\int\left(e^{9 x)^{1 / 3}}\left\{27+e^{3 x}\right\}^{1 / 3} d x\right.$
$=\int e^{3 x}\left(27+e^{3 x}\right)^{1 / 3} d x $
Put $t=27+e^{3 x} $
$\Rightarrow \, d t=3 e^{3 x} \,d x$
$\Rightarrow \, I=\frac{1}{3} \int t^{1 / 3} d t=\frac{1}{3}\left[t^{4 / 3}\right] \times \frac{3}{4}+C$
$\Rightarrow \, I=\frac{1}{4}\left(27+e^{3 x}\right)^{4 / 3}+C$