What is the integral of (x^2+e^{1-x}) ?

The integral of (x^2+e^{1-x}) is (x^3)/3-e^{1-x}+C

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integral of (x^2+e^{1-x})

\int (x^{2} + e^{1 - x}) d x

\frac{x ^{3}}{3} - e^{1 - x} + C

Solution steps

\int x^{2} + e^{1 - x} d x

Simplify

e^{1 - x} : e e^{- x}

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

\int x^{2} d x = \frac{x ^{3}}{3}

\int e e^{- x} d x = - e^{1 - x}

= \frac{x ^{3}}{3} - e^{1 - x}

Add a constant to the solution

= \frac{x ^{3}}{3} - e^{1 - x} + C

What is the integral of (x^2+e^{1-x}) ?
The integral of (x^2+e^{1-x}) is (x^3)/3-e^{1-x}+C