What is the integral of xe^{(4-x^2)} ?

The integral of xe^{(4-x^2)} is -1/2 e^{4-x^2}+C

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integral of xe^{(4-x^2)}

\int x e^{(4 - x^{2})} d x

- \frac{1}{2} e^{4 - x^{2}} + C

Solution steps

\int x e^{4 - x^{2}} d x

Apply u-substitution

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

= \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

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

Use the common integral:

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

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

Substitute back

u = 4 - x^{2}

= - \frac{1}{2} e^{4 - x^{2}}

Add a constant to the solution

= - \frac{1}{2} e^{4 - x^{2}} + C

\int 34 x d x

\frac{\partial}{\partial x} (7 e^{x^{5} y})

\int_{- 1}^{1} \frac{1}{x ^{2} - 2 x} d x

n = 1 \sum \infty \frac{2}{n} - \frac{2}{n + 1}

\int \frac{x ^{2}}{( 1 - x ^{2} ) ^{\frac{5}{2}}} d x

What is the integral of xe^{(4-x^2)} ?
The integral of xe^{(4-x^2)} is -1/2 e^{4-x^2}+C