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

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

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integral of 2xe^{4x}

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

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

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

= 2 \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

= 2 \cdot \frac{1}{16} \cdot \int e^{u} u d u

Apply Integration By Parts

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

= 2 \cdot \frac{1}{16} (e^{u} u - \int e^{u} d u)

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

= 2 \cdot \frac{1}{16} (e^{u} u - e^{u})

Substitute back

u = 4 x

= 2 \cdot \frac{1}{16} (e^{4 x} \cdot 4 x - e^{4 x})

Simplify

2 \cdot \frac{1}{16} (e^{4 x} \cdot 4 x - e^{4 x}) : \frac{1}{8} (4 e^{4 x} x - e^{4 x})

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

Add a constant to the solution

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

y^{^{'}} + 3 (tan (3 x)) y = - 10 cos (3 x)

\int_{0}^{2} \int 1

\int \frac{6 x ^{3} - 12 x ^{2} + 5}{x ^{2} - 2 x} d x

x \to \infty lim (e^{- 1})

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