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

The integral of xe^{-4x^2} is -1/8 e^{-4x^2}+C

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integral of xe^{-4x^2}

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

- \frac{1}{8} e^{- 4 x^{2}} + C

Solution steps

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

Apply u-substitution

= \int - \frac{e ^{u}}{8} d u

Take the constant out:

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

= - \frac{1}{8} \cdot \int e^{u} d u

Use the common integral:

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

= - \frac{1}{8} e^{u}

Substitute back

u = - 4 x^{2}

= - \frac{1}{8} e^{- 4 x^{2}}

Add a constant to the solution

= - \frac{1}{8} e^{- 4 x^{2}} + C

f (x) = (sin (x))^{x}

\int e^{2 x} d x

\frac{\partial}{\partial y} (\frac{x ^{2}}{4 y ^{4}})

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

\int \frac{1}{z ^{2} + 4} d z

What is the integral of xe^{-4x^2} ?
The integral of xe^{-4x^2} is -1/8 e^{-4x^2}+C