What is the integral of (2x+1)e^x ?

The integral of (2x+1)e^x is 2e^xx-e^x+C

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integral of (2x+1)e^x

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

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

Solution steps

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

Apply Integration By Parts

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

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

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

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

Simplify

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

= 2 e^{x} x - e^{x}

Add a constant to the solution

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

\frac{\partial}{\partial x} (6 x + 3 y)

\frac{d}{d x} (ln (x^{2} - 3))

y = 2 x^{4} - 16 x^{2} + 9, at x = - 4

y = \frac{2}{x}, (5, \frac{2}{5})

\int - 2 x^{- 3} d x