What is the integral of 2xe^{2x+1} ?

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

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integral of 2xe^{2x+1}

\int 2 x e^{2 x + 1} d x

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

Solution steps

\int 2 x e^{2 x + 1} 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 + 1} d x

Apply Integration By Parts

: \frac{1}{2} e^{2 x + 1} x - \int \frac{1}{2} e^{2 x + 1} d x

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

\int \frac{1}{2} e^{2 x + 1} d x = \frac{1}{4} e^{2 x + 1}

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

Simplify

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

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

Add a constant to the solution

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

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