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

= 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

= 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_{1}^{2} \frac{x ^{5} + 5 x ^{8}}{x ^{4}} d x

\frac{d}{d x} (x^{\frac{1}{3}} + 2 x)

\int_{0}^{9} x d x

y^{^{''}} + 4 y = t - 2 sin (2 t)

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

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