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$\int\left(27 \mathrm{e}^{9 \mathrm{x}}+\mathrm{e}^{12 \mathrm{x}}\right)^{1 / 3} \mathrm{dx}$ is equal to
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Verified Answer
The correct answer is:
$(1 / 4)\left(27+e^{3 x}\right)^{4 / 3}+C$
$$
\begin{array}{l}
\text { Let } \mathrm{I}=\int\left(27 \mathrm{e}^{9 \mathrm{x}}+\mathrm{e}^{12 \mathrm{x}}\right)^{1 / 3} \mathrm{dx} \\
=\int \mathrm{e}^{3 \mathrm{x}}\left(27+\mathrm{e}^{3 \mathrm{x}}\right)^{1 / 3} \mathrm{dx}
\end{array}
$$
Put $27+e^{3 x}=t \Rightarrow 3 e^{3 x} d x=d t$
$$
\begin{array}{l}
\therefore \mathrm{I}=\frac{1}{3} \int \mathrm{t}^{1 / 3} \mathrm{dt}=\frac{1}{3} \times \frac{\mathrm{t}^{4 / 3}}{4 / 3}+\mathrm{C} \\
=\frac{1}{4}\left(27+\mathrm{e}^{3 \mathrm{x}}\right)^{4 / 3}+\mathrm{C}
\end{array}
$$
\begin{array}{l}
\text { Let } \mathrm{I}=\int\left(27 \mathrm{e}^{9 \mathrm{x}}+\mathrm{e}^{12 \mathrm{x}}\right)^{1 / 3} \mathrm{dx} \\
=\int \mathrm{e}^{3 \mathrm{x}}\left(27+\mathrm{e}^{3 \mathrm{x}}\right)^{1 / 3} \mathrm{dx}
\end{array}
$$
Put $27+e^{3 x}=t \Rightarrow 3 e^{3 x} d x=d t$
$$
\begin{array}{l}
\therefore \mathrm{I}=\frac{1}{3} \int \mathrm{t}^{1 / 3} \mathrm{dt}=\frac{1}{3} \times \frac{\mathrm{t}^{4 / 3}}{4 / 3}+\mathrm{C} \\
=\frac{1}{4}\left(27+\mathrm{e}^{3 \mathrm{x}}\right)^{4 / 3}+\mathrm{C}
\end{array}
$$
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