What is the integral of (7x^3-6x+4)/(x^3-x^2) ?

The integral of (7x^3-6x+4)/(x^3-x^2) is (7x^2)/(x-1)+5ln|x-1|-7/(x-1)+2ln|x|+4/x+C

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integral of (7x^3-6x+4)/(x^3-x^2)

\int \frac{7 x ^{3} - 6 x + 4}{x ^{3} - x ^{2}} d x

\frac{7 x ^{2}}{x - 1} + 5 ln ∣ x - 1 ∣ - \frac{7}{x - 1} + 2 ln ∣ x ∣ + \frac{4}{x} + C

Solution steps

\int \frac{7 x ^{3} - 6 x + 4}{x ^{3} - x ^{2}} d x

Expand

\frac{7 x ^{3} - 6 x + 4}{x ^{3} - x ^{2}} : \frac{7 x ^{3}}{x ^{3} - x ^{2}} - \frac{6 x}{x ^{3} - x ^{2}} + \frac{4}{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{7 x ^{3}}{x ^{3} - x ^{2}} d x - \int \frac{6 x}{x ^{3} - x ^{2}} d x + \int \frac{4}{x ^{3} - x ^{2}} d x

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

\int \frac{6 x}{x ^{3} - x ^{2}} d x = 6 (- ln ∣ x ∣ + ln ∣ x - 1 ∣)

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

= 7 (\frac{x ^{2}}{x - 1} + ln ∣ x - 1 ∣ - \frac{1}{x - 1}) - 6 (- ln ∣ x ∣ + ln ∣ x - 1 ∣) + 4 (- ln ∣ x ∣ + \frac{1}{x} + ln ∣ x - 1 ∣)

Simplify

7 (\frac{x ^{2}}{x - 1} + ln ∣ x - 1 ∣ - \frac{1}{x - 1}) - 6 (- ln ∣ x ∣ + ln ∣ x - 1 ∣) + 4 (- ln ∣ x ∣ + \frac{1}{x} + ln ∣ x - 1 ∣) : \frac{7 x ^{2}}{x - 1} + 5 ln ∣ x - 1 ∣ - \frac{7}{x - 1} + 2 ln ∣ x ∣ + \frac{4}{x}

= \frac{7 x ^{2}}{x - 1} + 5 ln ∣ x - 1 ∣ - \frac{7}{x - 1} + 2 ln ∣ x ∣ + \frac{4}{x}

Add a constant to the solution

= \frac{7 x ^{2}}{x - 1} + 5 ln ∣ x - 1 ∣ - \frac{7}{x - 1} + 2 ln ∣ x ∣ + \frac{4}{x} + C

\int \frac{ln ^{2} ( x + 1 )}{x + 1} d x

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\int \frac{x ^{9}}{( x ^{4} - 1 ) ^{2}} d x

\int \frac{1}{x 3 - ln ^{2} ( x )} d x

What is the integral of (7x^3-6x+4)/(x^3-x^2) ?
The integral of (7x^3-6x+4)/(x^3-x^2) is (7x^2)/(x-1)+5ln|x-1|-7/(x-1)+2ln|x|+4/x+C