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

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

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integral of (2xe^{2x})

\int (2 x e^{2 x}) d x

e^{2 x} x - \frac{1}{2} e^{2 x} + C

Solution steps

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

Apply u-substitution

: \int \frac{e ^{u} u}{4} d u

= 2 \cdot \int \frac{e ^{u} u}{4} d u

Take the constant out:

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

= 2 \cdot \frac{1}{4} \cdot \int e^{u} u d u

Apply Integration By Parts

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

= 2 \cdot \frac{1}{4} (e^{u} u - \int e^{u} d u)

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

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

Substitute back

u = 2 x

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

Simplify

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

= e^{2 x} x - \frac{1}{2} e^{2 x}

Add a constant to the solution

= e^{2 x} x - \frac{1}{2} e^{2 x} + C

\int e^{7 x} d x

y^{^{''}} + 2 y^{^{'}} + 1 y = 1, y (1) = 1, y^{^{'}} (1) = 0

n \to \infty lim (n^{2} x)

f (x) = e^{13}

y^{^{'}} - \frac{4}{x} y = \frac{y ^{4}}{x ^{4}}, y (1) = 1

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