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

The integral of (x^2+2x-2)/(x^4-x^3) is -ln|x|+ln|x-1|-1/(x^2)+C

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

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

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

Solution steps

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

Expand

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

Apply the Sum Rule:

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

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

\int \frac{x ^{2}}{x ^{4} - x ^{3}} d x = - ln ∣ x ∣ + ln ∣ x - 1 ∣

\int \frac{2 x}{x ^{4} - x ^{3}} d x = 2 (- ln ∣ x ∣ + \frac{1}{x} + ln ∣ x - 1 ∣)

\int \frac{2}{x ^{4} - x ^{3}} d x = - 2 ln ∣ x ∣ + \frac{2}{x} + \frac{1}{x ^{2}} + 2 ln ∣ x - 1 ∣

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

Simplify

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

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

Add a constant to the solution

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

\frac{\partial}{\partial x} (x^{7} + x y^{9} + 1)

\frac{d}{d x} (27 arctan (x))

\frac{\partial}{\partial x} (x^{2} + x y + 3 x + 2 y + 5)

y^{^{''}} + 4 y^{^{'}} + 4 y = - 3 e^{2 t}, y (0) = 1, y^{^{'}} (0) = 0

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

What is the integral of (x^2+2x-2)/(x^4-x^3) ?
The integral of (x^2+2x-2)/(x^4-x^3) is -ln|x|+ln|x-1|-1/(x^2)+C