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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

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

- \frac{1}{v + 2} + C

Solution steps

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

Apply u-substitution

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

Apply the Power Rule

= - \frac{1}{u}

Substitute back

u = v + 2

= - \frac{1}{v + 2}

Add a constant to the solution

= - \frac{1}{v + 2} + C

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

\int (5 x^{4} + 6 x + 7) d x

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

\int x \cdot x^{'} d x

\int \frac{1}{x 5 + ln ( 3 x )} d x