What is the integral of 9xe^{6x^2} ?

The integral of 9xe^{6x^2} is 3/4 e^{6x^2}+C

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integral of 9xe^{6x^2}

\int 9 x e^{6 x^{2}} d x

\frac{3}{4} e^{6 x^{2}} + C

Solution steps

\int 9 x e^{6 x^{2}} d x

Take the constant out:

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

= 9 \cdot \int x e^{6 x^{2}} d x

Apply u-substitution

: \int \frac{e ^{u}}{12} d u

= 9 \cdot \int \frac{e ^{u}}{12} d u

Take the constant out:

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

= 9 \cdot \frac{1}{12} \cdot \int e^{u} d u

Use the common integral:

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

= 9 \cdot \frac{1}{12} e^{u}

Substitute back

u = 6 x^{2}

= 9 \cdot \frac{1}{12} e^{6 x^{2}}

Simplify

9 \cdot \frac{1}{12} e^{6 x^{2}} : \frac{3}{4} e^{6 x^{2}}

= \frac{3}{4} e^{6 x^{2}}

Add a constant to the solution

= \frac{3}{4} e^{6 x^{2}} + C

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What is the integral of 9xe^{6x^2} ?
The integral of 9xe^{6x^2} is 3/4 e^{6x^2}+C