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) (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

x \to 1 + lim ((x)^{\frac{1}{x - 1}})

(x - 1)^{2} + 2, (- 1, 2)

\frac{\partial}{\partial x} (- \frac{y}{y ^{2} + x ^{2}})

\frac{x}{4 + x ^{2}}

y^{^{'}} + y = x y^{3}

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