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

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

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

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

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

Solution steps

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

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

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

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

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

Add a constant to the solution

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

\frac{d ^{2}}{d x ^{2}} (x 1 - x^{2})

\int_{0}^{r} x r^{2} - x^{2} d x

\int \frac{x}{( x ^{2} - 4 ) ( x + 3 )} d x

y = 1 - e^{x}

\int 1 5^{x} d x

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