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

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

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

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

- ln ∣ x ∣ + \frac{2}{x} + ln ∣ x - 2 ∣ + C

Solution steps

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

Take the constant out:

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

= 4 \cdot \int \frac{1}{x ^{3} - 2 x ^{2}} d x

Take the partial fraction of

\frac{1}{x ^{3} - 2 x ^{2}} : - \frac{1}{4 x} - \frac{1}{2 x ^{2}} + \frac{1}{4 ( x - 2 )}

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

Apply the Sum Rule:

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

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

\int \frac{1}{4 x} d x = \frac{1}{4} ln ∣ x ∣

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

\int \frac{1}{4 ( x - 2 )} d x = \frac{1}{4} ln ∣ x - 2 ∣

= 4 (- \frac{1}{4} ln ∣ x ∣ - (- \frac{1}{2 x}) + \frac{1}{4} ln ∣ x - 2 ∣)

Simplify

4 (- \frac{1}{4} ln ∣ x ∣ - (- \frac{1}{2 x}) + \frac{1}{4} ln ∣ x - 2 ∣) : - ln ∣ x ∣ + \frac{2}{x} + ln ∣ x - 2 ∣

= - ln ∣ x ∣ + \frac{2}{x} + ln ∣ x - 2 ∣

Add a constant to the solution

= - ln ∣ x ∣ + \frac{2}{x} + ln ∣ x - 2 ∣ + C

x y^{^{'}} - 2 y = 6 x^{2}

\int f (x, y, z) d x

\int \frac{1}{100 - x ^{2}} d x

\frac{\partial}{\partial x} (\frac{- y ^{2}}{( x - y ) ^{2}})

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