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

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

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

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

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

Solution steps

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

Apply Integration By Parts

: \frac{x ^{3} + x ^{2} + x - 2}{x ( x - 1 )} - \int - \frac{2 x ^{3} + 2 x ^{2} - 7 x + 4}{x ^{2} ( x - 1 ) ^{2}} d x

= \frac{x ^{3} + x ^{2} + x - 2}{x ( x - 1 )} - \int - \frac{2 x ^{3} + 2 x ^{2} - 7 x + 4}{x ^{2} ( x - 1 ) ^{2}} d x

\int - \frac{2 x ^{3} + 2 x ^{2} - 7 x + 4}{x ^{2} ( x - 1 ) ^{2}} d x = - ln ∣ x - 1 ∣ + \frac{1}{x - 1} - ln ∣ x ∣ + \frac{4}{x}

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

Simplify

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

Add a constant to the solution

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

4 - x^{2}, \frac{1}{4} x^{3} - 2, 1, 2

\frac{d y}{d x} = \frac{8 x ^{2}}{9 + x ^{3}}

h^{2} + 7 h + 40

y^{^{'}} = x y^{8}

\int_{0}^{x} x d x

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