What is the integral of-(e^x(x+1))/3 ?

The integral of-(e^x(x+1))/3 is -1/3 e^xx+C

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integral of-(e^x(x+1))/3

\int - \frac{e ^{x} ( x + 1 )}{3} d x

- \frac{1}{3} e^{x} x + C

Solution steps

\int - \frac{e ^{x} ( x + 1 )}{3} d x

Take the constant out:

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

= - \frac{1}{3} \cdot \int e^{x} (x + 1) d x

Apply Integration By Parts

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

= - \frac{1}{3} (e^{x} (x + 1) - \int e^{x} d x)

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

= - \frac{1}{3} (e^{x} (x + 1) - e^{x})

Simplify

- \frac{1}{3} (e^{x} (x + 1) - e^{x}) : - \frac{1}{3} e^{x} x

= - \frac{1}{3} e^{x} x

Add a constant to the solution

= - \frac{1}{3} e^{x} x + C

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What is the integral of-(e^x(x+1))/3 ?
The integral of-(e^x(x+1))/3 is -1/3 e^xx+C