What is the integral of 8xe^{4x^2} ?

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

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integral of 8xe^{4x^2}

\int 8 x e^{4 x^{2}} d x

e^{4 x^{2}} + C

Solution steps

\int 8 x e^{4 x^{2}} d x

Take the constant out:

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

= 8 \cdot \int x e^{4 x^{2}} d x

Apply u-substitution

: \int \frac{e ^{u}}{8} d u

= 8 \cdot \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

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

Use the common integral:

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

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

Substitute back

u = 4 x^{2}

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

Simplify

8 \cdot \frac{1}{8} e^{4 x^{2}} : e^{4 x^{2}}

= e^{4 x^{2}}

Add a constant to the solution

= e^{4 x^{2}} + C

sin (x), \frac{π}{5}

f (x) = 4^{2}

\int ln (v) d v

\frac{\partial}{\partial x} (e^{- x^{3}})

f (x) = \frac{3}{x}