What is the integral of 16xe^{4x} ?

The integral of 16xe^{4x} is 4e^{4x}x-e^{4x}+C

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integral of 16xe^{4x}

\int 16 x e^{4 x} d x

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

Solution steps

\int 16 x e^{4 x} d x

Take the constant out:

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

= 16 \cdot \int x e^{4 x} d x

Apply u-substitution

: \int \frac{e ^{u} u}{16} d u

= 16 \cdot \int \frac{e ^{u} u}{16} d u

Take the constant out:

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

= 16 \cdot \frac{1}{16} \cdot \int e^{u} u d u

Apply Integration By Parts

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

= 16 \cdot \frac{1}{16} (e^{u} u - \int e^{u} d u)

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

= 16 \cdot \frac{1}{16} (e^{u} u - e^{u})

Substitute back

u = 4 x

= 16 \cdot \frac{1}{16} (e^{4 x} \cdot 4 x - e^{4 x})

Simplify

16 \cdot \frac{1}{16} (e^{4 x} \cdot 4 x - e^{4 x}) : 4 e^{4 x} x - e^{4 x}

= 4 e^{4 x} x - e^{4 x}

Add a constant to the solution

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

x^{2} + 16

3 x^{3} - 12 x^{2} + 15 x

f (t) = t^{2}

\int 8 y^{- 3} d y

f (x) = \frac{8}{x ^{2} + 4}, at x = 4

What is the integral of 16xe^{4x} ?
The integral of 16xe^{4x} is 4e^{4x}x-e^{4x}+C