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

The integral of 4xe^{4x^2} is 1/2 e^{4x^2}+C

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integral of 4xe^{4x^2}

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

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

Solution steps

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

Take the constant out:

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

= 4 \cdot \int x e^{4 x^{2}} d x

Apply u-substitution

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

= 4 \cdot \int \frac{e ^{u}}{8} d u

Take the constant out:

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

= 4 \cdot \frac{1}{8} \cdot \int e^{u} d u

Use the common integral:

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

= 4 \cdot \frac{1}{8} e^{u}

Substitute back

u = 4 x^{2}

= 4 \cdot \frac{1}{8} e^{4 x^{2}}

Simplify

4 \cdot \frac{1}{8} e^{4 x^{2}} : \frac{1}{2} e^{4 x^{2}}

= \frac{1}{2} e^{4 x^{2}}

Add a constant to the solution

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

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What is the integral of 4xe^{4x^2} ?
The integral of 4xe^{4x^2} is 1/2 e^{4x^2}+C