What is the integral of x/((4+xy)^2) ?

The integral of x/((4+xy)^2) is -1/(4+xy)+C

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integral of x/((4+xy)^2)

\int \frac{x}{( 4 + x y ) ^{2}} d y

- \frac{1}{4 + x y} + C

Solution steps

\int \frac{x}{( 4 + x y ) ^{2}} d y

Take the constant out:

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

= x \cdot \int \frac{1}{( 4 + x y ) ^{2}} d y

Apply u-substitution

: \int \frac{1}{x u ^{2}} d u

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

Take the constant out:

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

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

Apply the Power Rule

: - \frac{1}{u}

= x \frac{1}{x} (- \frac{1}{u})

Substitute back

u = 4 + x y

= x \frac{1}{x} (- \frac{1}{4 + x y})

Simplify

x \frac{1}{x} (- \frac{1}{4 + x y}) : - \frac{1}{4 + x y}

= - \frac{1}{4 + x y}

Add a constant to the solution

= - \frac{1}{4 + x y} + C

\int cos^{5} (9 - 11 x) d x

\frac{d}{d x} (\frac{3 x - 4}{2 x - 8})

\int 8 x - 5 d x

\frac{1}{3 ( 1 + x ^{2} )}, \frac{1}{6} x^{2}

x y^{^{''}} + 2 y^{^{'}} = 4 x^{2}

What is the integral of x/((4+xy)^2) ?
The integral of x/((4+xy)^2) is -1/(4+xy)+C