What is the integral of y/(3+xy) ?

The integral of y/(3+xy) is ln|3+yx|+C

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integral of y/(3+xy)

\int \frac{y}{3 + x y} d x

ln ∣ 3 + y x ∣ + C

Solution steps

\int \frac{y}{3 + x y} 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}{3 + y x} d x

Apply u-substitution

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

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

Take the constant out:

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= y \frac{1}{y} ln ∣ u ∣

Substitute back

u = 3 + y x

= y \frac{1}{y} ln ∣ 3 + y x ∣

Simplify

y \frac{1}{y} ln ∣ 3 + y x ∣ : ln ∣ 3 + y x ∣

= ln ∣ 3 + y x ∣

Add a constant to the solution

= ln ∣ 3 + y x ∣ + C

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\int_{0}^{1} 8 x d x

n = 0 \sum \infty \frac{n !}{2 n !}