What is the integral of (x^3)/3-e^{-x}+1 ?

The integral of (x^3)/3-e^{-x}+1 is (x^4)/(12)+e^{-x}+x+C

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integral of (x^3)/3-e^{-x}+1

\int \frac{x ^{3}}{3} - e^{- x} + 1 d x

\frac{x ^{4}}{12} + e^{- x} + x + C

Solution steps

\int \frac{x ^{3}}{3} - e^{- x} + 1 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{x ^{3}}{3} d x - \int e^{- x} d x + \int 1 d x

\int \frac{x ^{3}}{3} d x = \frac{x ^{4}}{12}

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

\int 1 d x = x

= \frac{x ^{4}}{12} - (- e^{- x}) + x

Simplify

= \frac{x ^{4}}{12} + e^{- x} + x

Add a constant to the solution

= \frac{x ^{4}}{12} + e^{- x} + x + C

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What is the integral of (x^3)/3-e^{-x}+1 ?
The integral of (x^3)/3-e^{-x}+1 is (x^4)/(12)+e^{-x}+x+C