Question

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

• A. $\left(1/4\right){\left(27+e^{3x}\right)}^{1/3}+C$
• B. $\left(1/4\right){\left(27+e^{3x}\right)}^{2/3}+C$
• C. $\left(1/3\right){\left(27+e^{3x}\right)}^{4/3}+C$
• D. $\left(1/4\right){\left(27+e^{3x}\right)}^{4/3}+C$
• E. $\left(3/4\right){\left(27+e^{3x}\right)}^{4/3}+C$

Answer 1

Answer: (D)

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