What is the integral of x(e^x+9) ?

The integral of x(e^x+9) is 9/2 x^2+e^xx-e^x+C

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integral of x(e^x+9)

\int x (e^{x} + 9) d x

\frac{9}{2} x^{2} + e^{x} x - e^{x} + C

Solution steps

\int x (e^{x} + 9) d x

Apply Integration By Parts

: \frac{1}{2} x^{2} (e^{x} + 9) - \int \frac{1}{2} e^{x} x^{2} d x

= \frac{1}{2} x^{2} (e^{x} + 9) - \int \frac{1}{2} e^{x} x^{2} d x

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

= \frac{1}{2} x^{2} (e^{x} + 9) - (\frac{1}{2} e^{x} x^{2} - e^{x} x + e^{x})

Simplify

\frac{1}{2} x^{2} (e^{x} + 9) - (\frac{1}{2} e^{x} x^{2} - e^{x} x + e^{x}) : \frac{9}{2} x^{2} + e^{x} x - e^{x}

= \frac{9}{2} x^{2} + e^{x} x - e^{x}

Add a constant to the solution

= \frac{9}{2} x^{2} + e^{x} x - e^{x} + C

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What is the integral of x(e^x+9) ?
The integral of x(e^x+9) is 9/2 x^2+e^xx-e^x+C