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

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

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integral of e^{2x}ln(e^x)

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

\frac{1}{2} e^{2 x} x - \frac{1}{4} e^{2 x} + C

Solution steps

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

Apply Integration By Parts

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

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

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

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

Add a constant to the solution

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

y = x^{2} - 8 x, at x = - 8

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