What is the integral of (e^x+1)*(x-4) ?

The integral of (e^x+1)*(x-4) is e^xx-5e^x+(x^2)/2-4x+C

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integral of (e^x+1)*(x-4)

\int (e^{x} + 1) \cdot (x - 4) d x

e^{x} x - 5 e^{x} + \frac{x ^{2}}{2} - 4 x + C

Solution steps

\int (e^{x} + 1) (x - 4) d x

Expand

(e^{x} + 1) (x - 4) : e^{x} x - 4 e^{x} + x - 4

= \int e^{x} x - 4 e^{x} + x - 4 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 4 e^{x} d x + \int x d x - \int 4 d x

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

\int 4 e^{x} d x = 4 e^{x}

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

\int 4 d x = 4 x

= e^{x} x - e^{x} - 4 e^{x} + \frac{x ^{2}}{2} - 4 x

Simplify

= e^{x} x - 5 e^{x} + \frac{x ^{2}}{2} - 4 x

Add a constant to the solution

= e^{x} x - 5 e^{x} + \frac{x ^{2}}{2} - 4 x + C

f (x) = 25 - x^{2}, (5, 0)

\frac{x + 7}{3 x ^{2} e ^{x}}

x \to 0 lim (ln (\frac{2}{x}))

y^{^{'}} = (x + y + 7)^{2}

What is the integral of (e^x+1)*(x-4) ?
The integral of (e^x+1)*(x-4) is e^xx-5e^x+(x^2)/2-4x+C