What is the integral of (6/5 x^5+4e^{-2x}) ?

The integral of (6/5 x^5+4e^{-2x}) is (x^6)/5-2e^{-2x}+C

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integral of (6/5 x^5+4e^{-2x})

\int (\frac{6}{5} x^{5} + 4 e^{- 2 x}) d x

\frac{x ^{6}}{5} - 2 e^{- 2 x} + C

Solution steps

\int \frac{6}{5} x^{5} + 4 e^{- 2 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{6}{5} x^{5} d x + \int 4 e^{- 2 x} d x

\int \frac{6}{5} x^{5} d x = \frac{x ^{6}}{5}

\int 4 e^{- 2 x} d x = - 2 e^{- 2 x}

= \frac{x ^{6}}{5} - 2 e^{- 2 x}

Add a constant to the solution

= \frac{x ^{6}}{5} - 2 e^{- 2 x} + C

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What is the integral of (6/5 x^5+4e^{-2x}) ?
The integral of (6/5 x^5+4e^{-2x}) is (x^6)/5-2e^{-2x}+C