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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

\int \frac{1}{- 4 x ^{2} - 20 x - 21} d x

\int \frac{1 + tan ( x )}{sin ( 2 x )} d x

\int \frac{3 x ^{4} + 2 x ^{2}}{x ^{2} + 1} d x

\int x cos (\frac{π}{4} x) d x

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