What is the integral of x^3e^{2x^2} ?

The integral of x^3e^{2x^2} is 1/8 (e^{2x^2}*2x^2-e^{2x^2})+C

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integral of x^3e^{2x^2}

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

\frac{1}{8} (e^{2 x^{2}} \cdot 2 x^{2} - e^{2 x^{2}}) + C

Solution steps

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

Apply u-substitution

: \int \frac{e ^{u} u}{8} d u

= \int \frac{e ^{u} u}{8} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{8} \cdot \int e^{u} u d u

Apply Integration By Parts

: e^{u} u - \int e^{u} d u

= \frac{1}{8} (e^{u} u - \int e^{u} d u)

\int e^{u} d u = e^{u}

= \frac{1}{8} (e^{u} u - e^{u})

Substitute back

u = 2 x^{2}

= \frac{1}{8} (e^{2 x^{2}} \cdot 2 x^{2} - e^{2 x^{2}})

Add a constant to the solution

= \frac{1}{8} (e^{2 x^{2}} \cdot 2 x^{2} - e^{2 x^{2}}) + C

\frac{d}{d x} (e^{s i n (x) + c o s (x)})

\frac{d}{d x} ((sin (20 x))^{3})

x + 3

\int_{- 1}^{- 4} \frac{1 - x ^{4}}{x ^{2}} d x

\frac{1}{x}

What is the integral of x^3e^{2x^2} ?
The integral of x^3e^{2x^2} is 1/8 (e^{2x^2}*2x^2-e^{2x^2})+C