What is the integral of (x+7)e^{0.1x+2} ?

The integral of (x+7)e^{0.1x+2} is 10e^{0.1x+2}x-30e^{0.1x+2}+C

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integral of (x+7)e^{0.1x+2}

\int (x + 7) e^{0.1 x + 2} d x

10 e^{0.1 x + 2} x - 30 e^{0.1 x + 2} + C

Solution steps

\int (x + 7) e^{0.1 x + 2} d x

Expand

(x + 7) e^{0.1 x + 2} : e^{0.1 x + 2} x + 7 e^{0.1 x + 2}

= \int e^{0.1 x + 2} x + 7 e^{0.1 x + 2} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int e^{0.1 x + 2} x d x + \int 7 e^{0.1 x + 2} d x

\int e^{0.1 x + 2} x d x = 10 e^{0.1 x + 2} x - 100 e^{0.1 x + 2}

\int 7 e^{0.1 x + 2} d x = 70 e^{0.1 x + 2}

= 10 e^{0.1 x + 2} x - 100 e^{0.1 x + 2} + 70 e^{0.1 x + 2}

Simplify

= 10 e^{0.1 x + 2} x - 30 e^{0.1 x + 2}

Add a constant to the solution

= 10 e^{0.1 x + 2} x - 30 e^{0.1 x + 2} + C

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What is the integral of (x+7)e^{0.1x+2} ?
The integral of (x+7)e^{0.1x+2} is 10e^{0.1x+2}x-30e^{0.1x+2}+C