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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

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

Simplify

e^{y^{2} + u} : e^{y^{2}} e^{u}

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

Use the common integral:

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

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

Substitute back

u = x^{2}

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

Simplify

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

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

Add a constant to the solution

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

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