Evaluate integral of (x^2)/(e^(x^3)) with respect to x

$\int \frac{x^{2}}{e^{x^{3}}} d x$

Factor $\frac{x^{2}}{e^{x^{3}}}$ .

$\int \frac{x^{2}}{e^{x^{3}}} d x$

Simplify the expression.

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Negate the exponent of $e^{x^{3}}$ and move it out of the denominator.

$\int x^{2} {(e^{x^{3}})}^{- 1} d x$

Multiply the exponents in ${(e^{x^{3}})}^{- 1}$ .

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Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$\int x^{2} e^{x^{3} \cdot - 1} d x$

Move $- 1$ to the left of $x^{3}$ .

$\int x^{2} e^{- 1 \cdot x^{3}} d x$

Rewrite $- 1 x^{3}$ as $- x^{3}$ .

$\int x^{2} e^{- x^{3}} d x$

Let $u_{2} = e^{- x^{3}}$ . Then $d u_{2} = - 3 \cdot x^{2} e^{- x^{3}} d x$ , so $- \frac{1}{3} d u_{2} = e^{- x^{3}} x^{2} d x$ . Rewrite using $u_{2}$ and $d$ $u_{2}$ .

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Let $u_{2} = e^{- x^{3}}$ . Find $\frac{d u_{2}}{d x}$ .

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Differentiate $e^{- x^{3}}$ .

$\frac{d}{d x} [e^{- x^{3}}]$

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = e^{x}$ and $g (x) = - x^{3}$ .

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To apply the Chain Rule, set $u_{1}$ as $- x^{3}$ .

$\frac{d}{d u_{1}} [e^{u_{1}}] \frac{d}{d x} [- x^{3}]$

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{1}} [a^{u_{1}}]$ is $a^{u_{1}} \ln (a)$ where $a$ = $e$ .

$e^{u_{1}} \frac{d}{d x} [- x^{3}]$

Replace all occurrences of $u_{1}$ with $- x^{3}$ .

$e^{- x^{3}} \frac{d}{d x} [- x^{3}]$

Since $- 1$ is constant with respect to $x$ , the derivative of $- x^{3}$ with respect to $x$ is $- \frac{d}{d x} [x^{3}]$ .

$e^{- x^{3}} (- \frac{d}{d x} [x^{3}])$

Move $- 1$ to the left of $e^{- x^{3}}$ .

$- 1 \cdot e^{- x^{3}} \frac{d}{d x} [x^{3}]$

Rewrite $- 1 e^{- x^{3}}$ as $- e^{- x^{3}}$ .

$- e^{- x^{3}} \frac{d}{d x} [x^{3}]$

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 3$ .

$- e^{- x^{3}} (3 x^{2})$

Multiply $3$ by $- 1$ .

$- 3 e^{- x^{3}} x^{2}$

Reorder $- 3 e^{- x^{3}}$ and $x^{2}$ .

$x^{2} (- 3 e^{- x^{3}})$

Reorder $x^{2}$ and $- 3$ .

$- 3 \cdot x^{2} e^{- x^{3}}$

Rewrite the problem using $u_{2}$ and $d u_{2}$ .

$\int \frac{1}{- 3} d u_{2}$

Move the negative in front of the fraction.

$\int - \frac{1}{3} d u_{2}$

Since $- \frac{1}{3}$ is constant with respect to $u_{2}$ , move $- \frac{1}{3}$ out of the integral.

$- \frac{1}{3} u_{2} + C$

Replace all occurrences of $u_{2}$ with $e^{- x^{3}}$ .

$- \frac{1}{3} e^{- x^{3}} + C$

Evaluate integral of (x^2)/(e^(x^3)) with respect to x

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