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

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

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

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

Solution steps

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

Factor

x^{2} - 1 : - (- x^{2} + 1)

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

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

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

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

\int 1 d x = x

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

Simplify

- (\frac{ln ∣ x + 1 ∣}{2} - \frac{ln ∣ x - 1 ∣}{2} - x) : - \frac{1}{2} ln ∣ x + 1 ∣ + \frac{1}{2} ln ∣ x - 1 ∣ + x

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

Add a constant to the solution

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

\int \frac{1}{9 + 2 x} d x

\int \frac{x}{x + x} d x

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

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

\int \frac{1}{2} sin (\frac{2 π}{5} x) d x

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