What is the integral of xe^{5x^2+9} ?

The integral of xe^{5x^2+9} is 1/10 e^{5x^2+9}+C

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integral of xe^{5x^2+9}

\int x e^{5 x^{2} + 9} d x

\frac{1}{10} e^{5 x^{2} + 9} + C

Solution steps

\int x e^{5 x^{2} + 9} d x

Apply u-substitution

: \int \frac{e ^{u}}{10} d u

= \int \frac{e ^{u}}{10} d u

Take the constant out:

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

= \frac{1}{10} \cdot \int e^{u} d u

Use the common integral:

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

= \frac{1}{10} e^{u}

Substitute back

u = 5 x^{2} + 9

= \frac{1}{10} e^{5 x^{2} + 9}

Add a constant to the solution

= \frac{1}{10} e^{5 x^{2} + 9} + C

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What is the integral of xe^{5x^2+9} ?
The integral of xe^{5x^2+9} is 1/10 e^{5x^2+9}+C