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

The integral of (-1)/(x+1^2) is -ln|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}}

- ln ∣ 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

Simplify

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

Apply u-substitution

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - ln ∣ u ∣

Substitute back

u = x + 1

= - ln ∣ x + 1 ∣

Add a constant to the solution

= - ln ∣ x + 1 ∣ + C

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

sin (- u + 3 t)

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

(x + 1) cos (x)

\frac{7}{x ^{2}}

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