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

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

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

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

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

Solution steps

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

Long division

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

= \int x + \frac{- x - 1}{x ^{3} - 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

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

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

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

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

Add a constant to the solution

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

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

ln (cos (2 x))

\int_{\frac{1}{3}}^{7} 10 x ln (3 x) d x

\int_{- 9}^{- 7} \frac{1}{x} d x

\int_{0}^{1} \frac{3}{x ( x + 1 )} d x

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