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

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

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integral of 1/((x+1))

integral

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

ln ∣ x + 1 ∣ + C

Solution steps

\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{1}{( x ^{3} + x ^{2} + x )}

\frac{cos ^{2} ( x )}{sin ^{2} ( x )}

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

\frac{1}{( x ^{2} + 3 )}

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