What is the integral of 3xe^{4x} ?

The integral of 3xe^{4x} is 3/16 (4e^{4x}x-e^{4x})+C

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integral of 3xe^{4x}

\int 3 x e^{4 x} d x

\frac{3}{16} (4 e^{4 x} x - e^{4 x}) + C

Solution steps

\int 3 x e^{4 x} d x

Take the constant out:

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

= 3 \cdot \int x e^{4 x} d x

Apply u-substitution

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

= 3 \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

= 3 \cdot \frac{1}{16} \cdot \int e^{u} u d u

Apply Integration By Parts

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

= 3 \cdot \frac{1}{16} (e^{u} u - \int e^{u} d u)

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

= 3 \cdot \frac{1}{16} (e^{u} u - e^{u})

Substitute back

u = 4 x

= 3 \cdot \frac{1}{16} (e^{4 x} \cdot 4 x - e^{4 x})

Simplify

3 \cdot \frac{1}{16} (e^{4 x} \cdot 4 x - e^{4 x}) : \frac{3}{16} (4 e^{4 x} x - e^{4 x})

= \frac{3}{16} (4 e^{4 x} x - e^{4 x})

Add a constant to the solution

= \frac{3}{16} (4 e^{4 x} x - e^{4 x}) + C

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

s^{2} + t^{2}

\int e^{c o s (π x)} sin (π x) d x

n = 0 \sum \infty (n) sin (\frac{1}{n})

y^{^{'}} + a y - b = 0

What is the integral of 3xe^{4x} ?
The integral of 3xe^{4x} is 3/16 (4e^{4x}x-e^{4x})+C