What is the integral of x^5e^{x^2} ?

The integral of x^5e^{x^2} is 1/2 (e^{x^2}x^4-2(e^{x^2}x^2-e^{x^2}))+C

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integral of x^5e^{x^2}

\int x^{5} e^{x^{2}} d x

\frac{1}{2} (e^{x^{2}} x^{4} - 2 (e^{x^{2}} x^{2} - e^{x^{2}})) + C

Solution steps

\int x^{5} e^{x^{2}} d x

Apply u-substitution

: \int \frac{e ^{u^{\frac{2}{5}}} u ^{\frac{1}{5}}}{5} d u

= \int \frac{e ^{u^{\frac{2}{5}}} u ^{\frac{1}{5}}}{5} d u

Take the constant out:

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

= \frac{1}{5} \cdot \int e^{u^{\frac{2}{5}}} u^{\frac{1}{5}} d u

Apply u-substitution

: \int \frac{5 e ^{v} v ^{2}}{2} d v

= \frac{1}{5} \cdot \int \frac{5 e ^{v} v ^{2}}{2} d v

Take the constant out:

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

= \frac{1}{5} \cdot \frac{5}{2} \cdot \int e^{v} v^{2} d v

Apply Integration By Parts

: v^{2} e^{v} - \int 2 v e^{v} d v

= \frac{1}{5} \cdot \frac{5}{2} (v^{2} e^{v} - \int 2 v e^{v} d v)

\int 2 v e^{v} d v = 2 (e^{v} v - e^{v})

= \frac{1}{5} \cdot \frac{5}{2} (v^{2} e^{v} - 2 (e^{v} v - e^{v}))

Substitute back

= \frac{1}{5} \cdot \frac{5}{2} (((x^{5})^{\frac{2}{5}})^{2} e^{(x^{5})^{\frac{2}{5}}} - 2 (e^{(x^{5})^{\frac{2}{5}}} (x^{5})^{\frac{2}{5}} - e^{(x^{5})^{\frac{2}{5}}}))

Simplify

\frac{1}{5} \cdot \frac{5}{2} (((x^{5})^{\frac{2}{5}})^{2} e^{(x^{5})^{\frac{2}{5}}} - 2 (e^{(x^{5})^{\frac{2}{5}}} (x^{5})^{\frac{2}{5}} - e^{(x^{5})^{\frac{2}{5}}})) : \frac{1}{2} (e^{x^{2}} x^{4} - 2 (e^{x^{2}} x^{2} - e^{x^{2}}))

= \frac{1}{2} (e^{x^{2}} x^{4} - 2 (e^{x^{2}} x^{2} - e^{x^{2}}))

Add a constant to the solution

= \frac{1}{2} (e^{x^{2}} x^{4} - 2 (e^{x^{2}} x^{2} - e^{x^{2}})) + C

\frac{d}{d x} (ln (cos (ln (x))))

\frac{d y}{d t} = 3 t e^{- y}

y = x arcsin (2 x)

x (\frac{d y}{d x}) = x^{3} + 16 x^{3} y

\frac{\partial}{\partial x} (3 x^{3} y - \frac{4}{x} + 2 y^{\frac{1}{2}})

What is the integral of x^5e^{x^2} ?
The integral of x^5e^{x^2} is 1/2 (e^{x^2}x^4-2(e^{x^2}x^2-e^{x^2}))+C