What is the integral of e^{4x}x ?

The integral of e^{4x}x is 1/16 (e^{4x}*4x-e^{4x})+C

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integral of e^{4x}x

\int e^{4 x} x d x

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

Solution steps

\int e^{4 x} x d x

Apply u-substitution

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

= \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

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

Apply Integration By Parts

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

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

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

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

Substitute back

u = 4 x

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

Add a constant to the solution

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

f (x) = 4 6 x^{2} + 7

\frac{\partial}{\partial x} (sin (x) + sin (y))

\int 4 x + x d x

5 x^{3} + 10 x^{2} - 2 x + 1

x^{2} - 2 x + 1

What is the integral of e^{4x}x ?
The integral of e^{4x}x is 1/16 (e^{4x}*4x-e^{4x})+C