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

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

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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

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

in t e g r a l \frac{( x + 1 )}{( x - 1 ) ^{2}}

in t e g r a l sin (- u + 3 t)

in t e g r a l - (\frac{( y ^{2} + 1 )}{( 2 x y )})

in t e g r a l \frac{2}{s ^{2}}

in t e g r a l \frac{1}{( x \cdot ( e ^{2} ) ^{x} )}

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