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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (u^2)/(u^2-1)

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

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

Solution steps

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

Factor

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

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

Take the constant out:

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

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

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

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

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}{- u ^{2} + 1} d u - \int 1 d u)

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

\int 1 d u = u

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

Simplify

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

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

Add a constant to the solution

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

\int (6 x^{2} - 4 x + 4) d x

\int \frac{x}{7 x ^{2} - 2} d x

\int x^{2} (5 + 2 x^{3}) d x

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

\int \frac{x ^{2}}{x ^{3} + 4 x ^{2} + 6 x} d x

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