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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of u/(1-u^2)

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

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

Solution steps

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

Apply u-substitution

: \int - \frac{1}{2 v} d v

= \int - \frac{1}{2 v} d v

Take the constant out:

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

= - \frac{1}{2} \cdot \int \frac{1}{v} d v

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= - \frac{1}{2} ln ∣ v ∣

Substitute back

v = 1 - u^{2}

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

Add a constant to the solution

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

\frac{d}{d x} (\frac{csc ^{3} ( x )}{ln ( x ^{2} )})

\frac{\partial}{\partial x} (y^{2} sin (4 x))

\frac{d}{d x} (2 ln (4 x))

\frac{d}{d x} (r x + s)

\int (2 x + 3)^{- 1} d x

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