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

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

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integral of y/(1+xy)

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

ln ∣ 1 + y x ∣ + C

Solution steps

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

= y \frac{1}{y} ln ∣ 1 + y x ∣

Simplify

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

= ln ∣ 1 + y x ∣

Add a constant to the solution

= ln ∣ 1 + y x ∣ + C

y = 1 - x^{2}, x = 2

\frac{x}{4 x + 8}

(x^{2} - 9) \frac{d y}{d x} + 6 y = (x + 3)^{2}

x \to 1 + lim (\frac{x - 4}{1 - x})