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

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

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integral of 1/(x(y+1))

\int \frac{1}{x ( y + 1 )} d y

\frac{1}{x} ln ∣ 1 + y ∣ + C

Solution steps

\int \frac{1}{x ( y + 1 )} d y

Take the constant out:

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

= \frac{1}{x} \cdot \int \frac{1}{1 + y} d y

Apply u-substitution

= \frac{1}{x} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{x} ln ∣ u ∣

Substitute back

u = 1 + y

= \frac{1}{x} ln ∣ 1 + y ∣

Add a constant to the solution

= \frac{1}{x} ln ∣ 1 + y ∣ + C

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