# Evaluate the following integrals. int x^2 6^{x^3+8}dx

Evaluate the following integrals.
$\int {x}^{2}{6}^{{x}^{3}+8}dx$
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Cullen
We have given
$\int {x}^{2}{6}^{{x}^{3}+8}dx$

$⇒3{x}^{2}dx=dt$
$⇒{x}^{2}dx=\frac{dt}{3}$
$\text{Then,}$
$\int {x}^{2}{6}^{{x}^{3}+8}=\frac{1}{3}\int {6}^{t}dt$
$=\frac{1}{3}\frac{{6}^{t}}{\mathrm{ln}\left(6\right)}$
$=\frac{1}{3}\cdot \frac{{6}^{{x}^{3}+8}}{\mathrm{ln}\left(6\right)}$
$=\frac{{6}^{{x}^{3}+8}}{3\mathrm{ln}\left(6\right)}$
$\text{Therefore,}$
$\int {x}^{2}{6}^{{x}^{3}+8}dx=\frac{{6}^{{x}^{3}+8}}{3\mathrm{ln}\left(6\right)}$