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

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

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

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

- \frac{1}{2} e^{- 2 x} + 2 (- e^{- x} x - e^{- x}) + \frac{x ^{3}}{3} + C

Solution steps

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

Apply u-substitution

: \int - e^{2 u} + 2 e^{u} u - u^{2} d u

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int e^{2 u} d u = \frac{1}{2} e^{2 u}

\int 2 e^{u} u d u = 2 (e^{u} u - e^{u})

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

= - \frac{1}{2} e^{2 u} + 2 (e^{u} u - e^{u}) - \frac{u ^{3}}{3}

Substitute back

u = - x

= - \frac{1}{2} e^{2 (- x)} + 2 (e^{- x} (- x) - e^{- x}) - \frac{( - x ) ^{3}}{3}

Simplify

- \frac{1}{2} e^{2 (- x)} + 2 (e^{- x} (- x) - e^{- x}) - \frac{( - x ) ^{3}}{3} : - \frac{1}{2} e^{- 2 x} + 2 (- e^{- x} x - e^{- x}) + \frac{x ^{3}}{3}

= - \frac{1}{2} e^{- 2 x} + 2 (- e^{- x} x - e^{- x}) + \frac{x ^{3}}{3}

Add a constant to the solution

= - \frac{1}{2} e^{- 2 x} + 2 (- e^{- x} x - e^{- x}) + \frac{x ^{3}}{3} + C

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