What is the integral of x/((xy+1)^2) ?

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

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integral of x/((xy+1)^2)

integral

\frac{x}{( x y + 1 ) ^{2}}

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

Solution steps

\int \frac{x}{( x y + 1 ) ^{2}} d x

Apply u-substitution

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

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

Expand

\frac{u - 1}{u ^{2}} : \frac{1}{u} - \frac{1}{u ^{2}}

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

\int \frac{1}{u} d u = ln ∣ u ∣

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

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

Substitute back

u = 1 + y x

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

Simplify

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

Add a constant to the solution

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

\frac{1}{( 2 ^{t} - 1 )}

\frac{( t - 1 ) ^{2}}{( 2 t + 1 ) ^{2}}

\frac{p}{( p ^{2} + 4 )}

4 x^{2} + 3 x + 2 \cdot 2 (x + 1)

(8 x - x^{4})^{3} (8 x^{3} - 16)

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