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$\int x^2 e^{x^3} d x$ equals
(a) $\frac{1}{3} e^{x^3}+C$
(b) $\frac{1}{3}+e^{x^2}+C$
(c) $\frac{1}{2} e^{x^3}+C$
(d) $\frac{1}{2} e^{x^2}+C$
$\int x^2 e^{x^3} d x$ equals
(a) $\frac{1}{3} e^{x^3}+C$
(b) $\frac{1}{3}+e^{x^2}+C$
(c) $\frac{1}{2} e^{x^3}+C$
(d) $\frac{1}{2} e^{x^2}+C$
Solution:
1950 Upvotes
Verified Answer
(a) Let $x^3=t \Rightarrow 3 x^2 d x=d t$
$\therefore \quad \int x^2 e^{x^3} d x=\frac{1}{3} \int e^t d t=\frac{1}{3} e^t+C=\frac{1}{3} e^{x^3}+C$
$\therefore \quad \int x^2 e^{x^3} d x=\frac{1}{3} \int e^t d t=\frac{1}{3} e^t+C=\frac{1}{3} e^{x^3}+C$
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