What is the integral of (4xy-2x^2)/(2x^3-x^2y) ?

The integral of (4xy-2x^2)/(2x^3-x^2y) is (4(-y+2x))/x-6ln|2x^3-x^2y|+C

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integral of (4xy-2x^2)/(2x^3-x^2y)

\int \frac{4 x y - 2 x ^{2}}{2 x ^{3} - x ^{2} y} d y

\frac{4 ( - y + 2 x )}{x} - 6 ln 2 x^{3} - x^{2} y + C

Solution steps

\int \frac{4 x y - 2 x ^{2}}{2 x ^{3} - x ^{2} y} d y

Apply u-substitution

: \int - \frac{2 ( - 2 u + 3 x ^{3} )}{x ^{3} u} d u

= \int - \frac{2 ( - 2 u + 3 x ^{3} )}{x ^{3} u} d u

Take the constant out:

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

= - \frac{2}{x ^{3}} \cdot \int \frac{- 2 u + 3 x ^{3}}{u} d u

Expand

\frac{- 2 u + 3 x ^{3}}{u} : - 2 + \frac{3 x ^{3}}{u}

= - \frac{2}{x ^{3}} \cdot \int - 2 + \frac{3 x ^{3}}{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{2}{x ^{3}} (- \int 2 d u + \int \frac{3 x ^{3}}{u} d u)

\int 2 d u = 2 u

\int \frac{3 x ^{3}}{u} d u = 3 x^{3} ln ∣ u ∣

= - \frac{2}{x ^{3}} (- 2 u + 3 x^{3} ln ∣ u ∣)

Substitute back

u = 2 x^{3} - x^{2} y

= - \frac{2}{x ^{3}} (- 2 (2 x^{3} - x^{2} y) + 3 x^{3} ln 2 x^{3} - x^{2} y)

Simplify

- \frac{2}{x ^{3}} (- 2 (2 x^{3} - x^{2} y) + 3 x^{3} ln 2 x^{3} - x^{2} y) : \frac{4 ( - y + 2 x )}{x} - 6 ln 2 x^{3} - x^{2} y

= \frac{4 ( - y + 2 x )}{x} - 6 ln 2 x^{3} - x^{2} y

Add a constant to the solution

= \frac{4 ( - y + 2 x )}{x} - 6 ln 2 x^{3} - x^{2} y + C

\int 90 (2 x - 5)^{8} d x

n = 5 \sum \infty 4 - \frac{4}{n ^{2}}

y^{^{'}} = \frac{2 t}{3 y ^{2}}

y^{^{'}} = 14 x y

\frac{d}{d x} (7 + 6 x)

What is the integral of (4xy-2x^2)/(2x^3-x^2y) ?
The integral of (4xy-2x^2)/(2x^3-x^2y) is (4(-y+2x))/x-6ln|2x^3-x^2y|+C