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

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

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integral of (1/x+x/e-x^e+e^x-2^{e+1})

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

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

Solution steps

\int \frac{1}{x} + \frac{x}{e} - x^{e} + e^{x} - 2^{e + 1} 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 \frac{1}{x} d x + \int \frac{x}{e} d x - \int x^{e} d x + \int e^{x} d x - \int 2^{e + 1} d x

\int \frac{1}{x} d x = ln ∣ x ∣

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

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

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

\int 2^{e + 1} d x = 2^{e + 1} x

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

Add a constant to the solution

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

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