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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

integral

\frac{1}{( x + 1 ) ^{2}}

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

Solution steps

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

Apply u-substitution

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

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

Apply the Power Rule

: - \frac{1}{u}

= - \frac{1}{u}

Substitute back

u = x + 1

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

Add a constant to the solution

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

\int x^{\frac{1}{2}} + 4 d x

\int \frac{2}{π} arcsin (x - 1) + 1 d x

\int x^{6} sin (6 x) d x

\int \frac{x ^{2}}{3} + 1 d x

\int e^{3 x} e^{2 x} d x