What is the integral of (x^e-x/e+2ex) ?

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

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integral of (x^e-x/e+2ex)

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

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

Solution steps

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

\int x^{e} d x = \frac{x ^{e + 1}}{e + 1}

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

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

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

Add a constant to the solution

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

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