What is the integral of (7x^3-7x+4)/(x^3-x^2) ?

The integral of (7x^3-7x+4)/(x^3-x^2) is (7x^3-7x+4)/(x(x-1))+4ln|x-1|-4/(x-1)+3ln|x|+8/x+C

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integral of (7x^3-7x+4)/(x^3-x^2)

\int \frac{7 x ^{3} - 7 x + 4}{x ^{3} - x ^{2}} d x

\frac{7 x ^{3} - 7 x + 4}{x ( x - 1 )} + 4 ln ∣ x - 1 ∣ - \frac{4}{x - 1} + 3 ln ∣ x ∣ + \frac{8}{x} + C

Solution steps

\int \frac{7 x ^{3} - 7 x + 4}{x ^{3} - x ^{2}} d x

Apply Integration By Parts

= \frac{7 x ^{3} - 7 x + 4}{x ( x - 1 )} - \int - \frac{7 x ^{3} - 14 x ^{2} + 19 x - 8}{x ^{2} ( x - 1 ) ^{2}} d x

\int - \frac{7 x ^{3} - 14 x ^{2} + 19 x - 8}{x ^{2} ( x - 1 ) ^{2}} d x = - 4 ln ∣ x - 1 ∣ + \frac{4}{x - 1} - 3 ln ∣ x ∣ - \frac{8}{x}

= \frac{7 x ^{3} - 7 x + 4}{x ( x - 1 )} - (- 4 ln ∣ x - 1 ∣ + \frac{4}{x - 1} - 3 ln ∣ x ∣ - \frac{8}{x})

Simplify

= \frac{7 x ^{3} - 7 x + 4}{x ( x - 1 )} + 4 ln ∣ x - 1 ∣ - \frac{4}{x - 1} + 3 ln ∣ x ∣ + \frac{8}{x}

Add a constant to the solution

= \frac{7 x ^{3} - 7 x + 4}{x ( x - 1 )} + 4 ln ∣ x - 1 ∣ - \frac{4}{x - 1} + 3 ln ∣ x ∣ + \frac{8}{x} + C

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What is the integral of (7x^3-7x+4)/(x^3-x^2) ?
The integral of (7x^3-7x+4)/(x^3-x^2) is (7x^3-7x+4)/(x(x-1))+4ln|x-1|-4/(x-1)+3ln|x|+8/x+C