What is the integral of (-8/(e^{-x)}+2ln(x)) ?

The integral of (-8/(e^{-x)}+2ln(x)) is -8e^x+2(xln(x)-x)+C

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integral of (-8/(e^{-x)}+2ln(x))

\int (- \frac{8}{e ^{- x}} + 2 ln (x)) d x

- 8 e^{x} + 2 (x ln (x) - x) + C

Solution steps

\int - \frac{8}{e ^{- x}} + 2 ln (x) 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 \frac{8}{e ^{- x}} d x + \int 2 ln (x) d x

\int \frac{8}{e ^{- x}} d x = 8 e^{x}

\int 2 ln (x) d x = 2 (x ln (x) - x)

= - 8 e^{x} + 2 (x ln (x) - x)

Add a constant to the solution

= - 8 e^{x} + 2 (x ln (x) - x) + C

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What is the integral of (-8/(e^{-x)}+2ln(x)) ?
The integral of (-8/(e^{-x)}+2ln(x)) is -8e^x+2(xln(x)-x)+C