What is the integral of (6xe^{x^2+1}) ?

The integral of (6xe^{x^2+1}) is 3e^{x^2+1}+C

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integral of (6xe^{x^2+1})

\int (6 x e^{x^{2} + 1}) d x

3 e^{x^{2} + 1} + C

Solution steps

\int 6 x e^{x^{2} + 1} d x

Take the constant out:

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

= 6 \cdot \int x e^{x^{2} + 1} d x

Apply u-substitution

: \int \frac{e ^{u}}{2} d u

= 6 \cdot \int \frac{e ^{u}}{2} d u

Take the constant out:

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

= 6 \cdot \frac{1}{2} \cdot \int e^{u} d u

Use the common integral:

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

= 6 \cdot \frac{1}{2} e^{u}

Substitute back

u = x^{2} + 1

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

Simplify

6 \cdot \frac{1}{2} e^{x^{2} + 1} : 3 e^{x^{2} + 1}

= 3 e^{x^{2} + 1}

Add a constant to the solution

= 3 e^{x^{2} + 1} + C

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What is the integral of (6xe^{x^2+1}) ?
The integral of (6xe^{x^2+1}) is 3e^{x^2+1}+C