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

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

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integral of (x^4-x+1)/(x^3-x^2)

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

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

Solution steps

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

Expand

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

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

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

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

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

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

Simplify

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

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

Add a constant to the solution

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

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