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

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

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integral of y/(x+1)

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

y ln ∣ 1 + x ∣ + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= y ln ∣ u ∣

Substitute back

u = 1 + x

= y ln ∣ 1 + x ∣

Add a constant to the solution

= y ln ∣ 1 + x ∣ + C

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