What is the integral of (e^x+2)(x-5) ?

The integral of (e^x+2)(x-5) is e^xx-6e^x+x^2-10x+C

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integral of (e^x+2)(x-5)

\int (e^{x} + 2) (x - 5) d x

e^{x} x - 6 e^{x} + x^{2} - 10 x + C

Solution steps

\int (e^{x} + 2) (x - 5) d x

Expand

(e^{x} + 2) (x - 5) : e^{x} x - 5 e^{x} + 2 x - 10

= \int e^{x} x - 5 e^{x} + 2 x - 10 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^{x} x d x - \int 5 e^{x} d x + \int 2 x d x - \int 10 d x

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

\int 5 e^{x} d x = 5 e^{x}

\int 2 x d x = x^{2}

\int 10 d x = 10 x

= e^{x} x - e^{x} - 5 e^{x} + x^{2} - 10 x

Simplify

= e^{x} x - 6 e^{x} + x^{2} - 10 x

Add a constant to the solution

= e^{x} x - 6 e^{x} + x^{2} - 10 x + C

\int \frac{sin ( 2 x )}{20 + cos ^{2} ( x )} d x

y = ln (x^{3} sin^{2} (x))

\int x^{7} x^{4} + 2 d x

\frac{1}{x + 5}

y = 34 (3 x + 1)^{- \frac{2}{5}}

What is the integral of (e^x+2)(x-5) ?
The integral of (e^x+2)(x-5) is e^xx-6e^x+x^2-10x+C