What is the integral of (25x^2-20x+2)/(x^3-5x^2+4x) ?

The integral of (25x^2-20x+2)/(x^3-5x^2+4x) is

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integral of (25x^2-20x+2)/(x^3-5x^2+4x)

\int \frac{25 x ^{2} - 20 x + 2}{x ^{3} - 5 x ^{2} + 4 x} d x

\frac{1}{2} ln ∣ x ∣ + 6 ln ∣ x - 1 ∣ - \frac{13}{2} ln ∣ x - 4 ∣ + \frac{25}{2} ln 4 x^{2} - 20 x + 16 - \frac{125}{6} (ln \frac{2}{3} x - \frac{2}{3} - ln \frac{2}{3} x - \frac{8}{3}) + C

Solution steps

\int \frac{25 x ^{2} - 20 x + 2}{x ^{3} - 5 x ^{2} + 4 x} d x

Expand

\frac{25 x ^{2} - 20 x + 2}{x ^{3} - 5 x ^{2} + 4 x} : \frac{25 x ^{2}}{x ^{3} - 5 x ^{2} + 4 x} - \frac{20 x}{x ^{3} - 5 x ^{2} + 4 x} + \frac{2}{x ^{3} - 5 x ^{2} + 4 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 \frac{25 x ^{2}}{x ^{3} - 5 x ^{2} + 4 x} d x - \int \frac{20 x}{x ^{3} - 5 x ^{2} + 4 x} d x + \int \frac{2}{x ^{3} - 5 x ^{2} + 4 x} d x

\int \frac{25 x ^{2}}{x ^{3} - 5 x ^{2} + 4 x} d x = 50 (\frac{1}{4} ln 4 x^{2} - 20 x + 16 - \frac{5}{12} (ln - \frac{2}{3} + \frac{2}{3} x - ln - \frac{8}{3} + \frac{2}{3} x))

\int \frac{20 x}{x ^{3} - 5 x ^{2} + 4 x} d x = 20 (- \frac{1}{3} ln ∣ x - 1 ∣ + \frac{1}{3} ln ∣ x - 4 ∣)

\int \frac{2}{x ^{3} - 5 x ^{2} + 4 x} d x = 2 (\frac{1}{4} ln ∣ x ∣ - \frac{1}{3} ln ∣ x - 1 ∣ + \frac{1}{12} ln ∣ x - 4 ∣)

= 50 (\frac{1}{4} ln 4 x^{2} - 20 x + 16 - \frac{5}{12} (ln - \frac{2}{3} + \frac{2}{3} x - ln - \frac{8}{3} + \frac{2}{3} x)) - 20 (- \frac{1}{3} ln ∣ x - 1 ∣ + \frac{1}{3} ln ∣ x - 4 ∣) + 2 (\frac{1}{4} ln ∣ x ∣ - \frac{1}{3} ln ∣ x - 1 ∣ + \frac{1}{12} ln ∣ x - 4 ∣)

Simplify

50 (\frac{1}{4} ln 4 x^{2} - 20 x + 16 - \frac{5}{12} (ln - \frac{2}{3} + \frac{2}{3} x - ln - \frac{8}{3} + \frac{2}{3} x)) - 20 (- \frac{1}{3} ln ∣ x - 1 ∣ + \frac{1}{3} ln ∣ x - 4 ∣) + 2 (\frac{1}{4} ln ∣ x ∣ - \frac{1}{3} ln ∣ x - 1 ∣ + \frac{1}{12} ln ∣ x - 4 ∣) : \frac{1}{2} ln ∣ x ∣ + 6 ln ∣ x - 1 ∣ - \frac{13}{2} ln ∣ x - 4 ∣ + \frac{25}{2} ln 4 x^{2} - 20 x + 16 - \frac{125}{6} (ln \frac{2}{3} x - \frac{2}{3} - ln \frac{2}{3} x - \frac{8}{3})

= \frac{1}{2} ln ∣ x ∣ + 6 ln ∣ x - 1 ∣ - \frac{13}{2} ln ∣ x - 4 ∣ + \frac{25}{2} ln 4 x^{2} - 20 x + 16 - \frac{125}{6} (ln \frac{2}{3} x - \frac{2}{3} - ln \frac{2}{3} x - \frac{8}{3})

Add a constant to the solution

= \frac{1}{2} ln ∣ x ∣ + 6 ln ∣ x - 1 ∣ - \frac{13}{2} ln ∣ x - 4 ∣ + \frac{25}{2} ln 4 x^{2} - 20 x + 16 - \frac{125}{6} (ln \frac{2}{3} x - \frac{2}{3} - ln \frac{2}{3} x - \frac{8}{3}) + C

\int e^{x} + \frac{1}{6 x} d x

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\int \frac{x + 3}{x - 1} d x

\int (1 + tan (θ))^{2} d θ

What is the integral of (25x^2-20x+2)/(x^3-5x^2+4x) ?
The integral of (25x^2-20x+2)/(x^3-5x^2+4x) is