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

The integral of 2xe^{x^2+1} is e^{x^2+1}+C

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integral of 2xe^{x^2+1}

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

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

Solution steps

\int 2 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

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

Apply u-substitution

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

= 2 \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

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

Use the common integral:

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

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

Substitute back

u = x^{2} + 1

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

Simplify

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

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

Add a constant to the solution

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

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