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

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

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

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

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

Solution steps

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

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

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

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

\int e d x = e x

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

Add a constant to the solution

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

\int sin (x) - x cos (x) d x

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

\int u u + 4 d u

\frac{1}{( 3 + t ) ^{2}}

\frac{d}{d x} ((x - 5) (5 x + 4))

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