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)

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

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

Solution steps

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

Take the constant out:

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

= - \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})

Simplify

= \frac{1}{x + 1}

Add a constant to the solution

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

\frac{\partial}{\partial x} (e^{- 8 x^{2} - 5 y^{2}})

n = 0 \sum \infty - \frac{1}{n + 1}

\int e^{3 x} (sin (3 x) + cos (3 x)) d x

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

n = 0 \sum \infty \frac{1}{3 n}

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