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

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

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

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

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

Solution steps

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

Expand

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

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

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

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

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

Add a constant to the solution

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

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