What is the integral of x^3e^{4x^2} ?

The integral of x^3e^{4x^2} is 1/32 (e^{4x^2}*4x^2-e^{4x^2})+C

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integral of x^3e^{4x^2}

\int x^{3} e^{4 x^{2}} d x

\frac{1}{32} (e^{4 x^{2}} \cdot 4 x^{2} - e^{4 x^{2}}) + C

Solution steps

\int x^{3} e^{4 x^{2}} d x

Apply u-substitution

: \int \frac{e ^{u} u}{32} d u

= \int \frac{e ^{u} u}{32} d u

Take the constant out:

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

= \frac{1}{32} \cdot \int e^{u} u d u

Apply Integration By Parts

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

= \frac{1}{32} (e^{u} u - \int e^{u} d u)

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

= \frac{1}{32} (e^{u} u - e^{u})

Substitute back

u = 4 x^{2}

= \frac{1}{32} (e^{4 x^{2}} \cdot 4 x^{2} - e^{4 x^{2}})

Add a constant to the solution

= \frac{1}{32} (e^{4 x^{2}} \cdot 4 x^{2} - e^{4 x^{2}}) + C

\frac{d}{d x} (\frac{2 sin ( x )}{sec ( x )})

\frac{2}{( s - 4 ) ^{2}}

f (x) = \frac{7}{x}, 7, at x = 5

\int_{0}^{4} 2 π x x d x

\frac{d}{d x} (\frac{1}{1 + e ^{\frac{1}{x}}})

What is the integral of x^3e^{4x^2} ?
The integral of x^3e^{4x^2} is 1/32 (e^{4x^2}*4x^2-e^{4x^2})+C