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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/((-x^2+1))

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

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

Solution steps

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

Use the common integral:

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

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

Add a constant to the solution

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

\int \frac{x}{x ^{2} + 8 ^{2}} d x

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

\int \frac{1}{5} e^{x} d x

\int (a x + b y) d x

\int \frac{y ^{2}}{( y + 1 )} d y

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