What is the integral of 10xe^{0.1x^2} ?

The integral of 10xe^{0.1x^2} is 50e^{0.1x^2}+C

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integral of 10xe^{0.1x^2}

\int 10 x e^{0.1 x^{2}} d x

50 e^{0.1 x^{2}} + C

Solution steps

\int 10 x e^{0.1 x^{2}} d x

Take the constant out:

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

= 10 \cdot \int x e^{0.1 x^{2}} d x

Apply u-substitution

= 10 \cdot \int \frac{e ^{u}}{0.2} d u

Take the constant out:

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

= 10 \cdot \frac{1}{0.2} \cdot \int e^{u} d u

Simplify

= 10 \cdot 5 \cdot \int e^{u} d u

Use the common integral:

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

= 10 \cdot 5 e^{u}

Substitute back

u = 0.1 x^{2}

= 10 \cdot 5 e^{0.1 x^{2}}

Simplify

= 50 e^{0.1 x^{2}}

Add a constant to the solution

= 50 e^{0.1 x^{2}} + C

What is the integral of 10xe^{0.1x^2} ?
The integral of 10xe^{0.1x^2} is 50e^{0.1x^2}+C