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

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

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integral of-e^x(x+1)

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

- e^{x} x + C

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Apply Integration By Parts

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

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

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

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

Expand

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

= - e^{x} x

Add a constant to the solution

= - e^{x} x + C

y = \frac{x ^{2} - 4}{x}

\frac{1}{s ^{2} + s ^{2}}

x \to 6 lim (cos (x \cdot π))

\frac{\partial}{\partial x} (\frac{I}{2 x})

\frac{\partial}{\partial x} (2 x^{2} e^{y})