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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of t/(1-t^2)

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

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

Solution steps

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

Apply u-substitution

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

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}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - \frac{1}{2} ln ∣ u ∣

Substitute back

u = 1 - t^{2}

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

Add a constant to the solution

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

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

\int \frac{ln ( x ^{2} )}{x ^{2}} d x

\frac{d}{d x} (arccos (\frac{7}{x}))

\int \frac{4}{3 x ^{2} 5 - 7 x ^{2}} d x

\frac{d}{d x} (\frac{4 x}{3 - cot ( x )})

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