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

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

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

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

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

Solution steps

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

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

\int \frac{1}{1 - x} d x = - ln ∣ 1 - x ∣

= \frac{1}{2} e^{2 x} - ln ∣ 1 - x ∣

Add a constant to the solution

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

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