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

The integral of 1/((v+1)^2) is -1/(v+1)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of 1/((v+1)^2)

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

- \frac{1}{v + 1} + C

Solution steps

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

Apply u-substitution

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

Apply the Power Rule

= - \frac{1}{u}

Substitute back

u = v + 1

= - \frac{1}{v + 1}

Add a constant to the solution

= - \frac{1}{v + 1} + C