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

The integral of x/(1+xy) is 1/(y^2)(1+yx-ln|1+yx|)+C

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integral of x/(1+xy)

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

\frac{1}{y ^{2}} (1 + y x - ln ∣ 1 + y x ∣) + C

Solution steps

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

Apply u-substitution

: \int \frac{u - 1}{y ^{2} u} d u

= \int \frac{u - 1}{y ^{2} u} d u

Take the constant out:

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

= \frac{1}{y ^{2}} \cdot \int \frac{u - 1}{u} d u

Expand

\frac{u - 1}{u} : 1 - \frac{1}{u}

= \frac{1}{y ^{2}} \cdot \int 1 - \frac{1}{u} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{y ^{2}} (\int 1 d u - \int \frac{1}{u} d u)

\int 1 d u = u

\int \frac{1}{u} d u = ln ∣ u ∣

= \frac{1}{y ^{2}} (u - ln ∣ u ∣)

Substitute back

u = 1 + y x

= \frac{1}{y ^{2}} (1 + y x - ln ∣ 1 + y x ∣)

Add a constant to the solution

= \frac{1}{y ^{2}} (1 + y x - ln ∣ 1 + y x ∣) + C

\frac{\partial}{\partial x} (\frac{3}{t + 1})

f (t)

f (x) = (1 + cos^{2} (x))^{7}

\int \frac{1}{2 sin ( θ )} d θ

\frac{\partial ^{2}}{\partial u ( v ) \partial v} (8 v e^{- 6 u (v)})

What is the integral of x/(1+xy) ?
The integral of x/(1+xy) is 1/(y^2)(1+yx-ln|1+yx|)+C