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

The integral of (7+2x)e^x is 2e^xx+5e^x+C

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integral of (7+2x)e^x

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

2 e^{x} x + 5 e^{x} + C

Solution steps

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

Apply Integration By Parts

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

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

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

= (7 + 2 x) e^{x} - 2 e^{x}

Simplify

(7 + 2 x) e^{x} - 2 e^{x} : 2 e^{x} x + 5 e^{x}

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

Add a constant to the solution

= 2 e^{x} x + 5 e^{x} + C

\int \frac{( x + 1 ) ^{2}}{x ^{2} + 1} d x

\int_{0}^{4} 1 + \frac{9 x}{4} d x

\int_{0}^{π} sin (4 x) d x

y = - 2 x^{2}, (- 5, 50)

y = \frac{1}{2} x^{2} + 1, (0, 1)