Mathematics | VITEEE 2019 | Discussion Page

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

$(1/4)(27+e^{3x})^{1/3}+C$

$(1/4)(27+e^{3x})^{2/3}+C$

$(1/3)(27+e^{3x})^{4/3}+C$

$(1/4)(27+e^{3x})^{4/3}+C$

Option D

Let I = $\int_{}^{} (27e^{9x}+e^{12x})^{1/3}dx$

= $\int_{}^{}e^{3x} (27+e^{3x})^{1/3}dx$

Put $27+e^{3x}$ = t 3 $e^{3x}$ dx = dt

I = $\frac{1}{3}\int_{}^{}t^{1/3}dt$

= $\frac{1}{3}\frac{t^{4/3}}{4/3}+C$

= $(1/4)(27+e^{3x})^{4/3}+C$

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