What is the integral of 5xe^{0.005x^2} ?

The integral of 5xe^{0.005x^2} is 500e^{0.005x^2}+C

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integral of 5xe^{0.005x^2}

\int 5 x e^{0.005 x^{2}} d x

500 e^{0.005 x^{2}} + C

Solution steps

\int 5 x e^{0.005 x^{2}} d x

Take the constant out:

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

= 5 \cdot \int x e^{0.005 x^{2}} d x

Apply u-substitution

: \int \frac{e ^{u}}{0.01} d u

= 5 \cdot \int \frac{e ^{u}}{0.01} d u

Take the constant out:

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

= 5 \cdot \frac{1}{0.01} \cdot \int e^{u} d u

Simplify

= 5 \cdot 100 \cdot \int e^{u} d u

Use the common integral:

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

= 5 \cdot 100 e^{u}

Substitute back

u = 0.005 x^{2}

= 5 \cdot 100 e^{0.005 x^{2}}

Simplify

= 500 e^{0.005 x^{2}}

Add a constant to the solution

= 500 e^{0.005 x^{2}} + C

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What is the integral of 5xe^{0.005x^2} ?
The integral of 5xe^{0.005x^2} is 500e^{0.005x^2}+C