What is the integral of (10x^2-8x-6)/(x^3-2x^2-3x) ?

The integral of (10x^2-8x-6)/(x^3-2x^2-3x) is 3ln|x+1|+5ln|x-3|+2ln|x|+C

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integral of (10x^2-8x-6)/(x^3-2x^2-3x)

\int \frac{10 x ^{2} - 8 x - 6}{x ^{3} - 2 x ^{2} - 3 x} d x

3 ln ∣ x + 1 ∣ + 5 ln ∣ x - 3 ∣ + 2 ln ∣ x ∣ + C

Solution steps

\int \frac{10 x ^{2} - 8 x - 6}{x ^{3} - 2 x ^{2} - 3 x} d x

Expand

\frac{10 x ^{2} - 8 x - 6}{x ^{3} - 2 x ^{2} - 3 x} : \frac{10 x ^{2}}{x ^{3} - 2 x ^{2} - 3 x} - \frac{8 x}{x ^{3} - 2 x ^{2} - 3 x} - \frac{6}{x ^{3} - 2 x ^{2} - 3 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{10 x ^{2}}{x ^{3} - 2 x ^{2} - 3 x} d x - \int \frac{8 x}{x ^{3} - 2 x ^{2} - 3 x} d x - \int \frac{6}{x ^{3} - 2 x ^{2} - 3 x} d x

\int \frac{10 x ^{2}}{x ^{3} - 2 x ^{2} - 3 x} d x = 10 (\frac{1}{4} ln ∣ x + 1 ∣ + \frac{3}{4} ln ∣ x - 3 ∣)

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

\int \frac{6}{x ^{3} - 2 x ^{2} - 3 x} d x = 6 (- \frac{1}{3} ln ∣ x ∣ + \frac{1}{4} ln ∣ x + 1 ∣ + \frac{1}{12} ln ∣ x - 3 ∣)

= 10 (\frac{1}{4} ln ∣ x + 1 ∣ + \frac{3}{4} ln ∣ x - 3 ∣) - 8 (- \frac{1}{4} ln ∣ x + 1 ∣ + \frac{1}{4} ln ∣ x - 3 ∣) - 6 (- \frac{1}{3} ln ∣ x ∣ + \frac{1}{4} ln ∣ x + 1 ∣ + \frac{1}{12} ln ∣ x - 3 ∣)

Simplify

10 (\frac{1}{4} ln ∣ x + 1 ∣ + \frac{3}{4} ln ∣ x - 3 ∣) - 8 (- \frac{1}{4} ln ∣ x + 1 ∣ + \frac{1}{4} ln ∣ x - 3 ∣) - 6 (- \frac{1}{3} ln ∣ x ∣ + \frac{1}{4} ln ∣ x + 1 ∣ + \frac{1}{12} ln ∣ x - 3 ∣) : 3 ln ∣ x + 1 ∣ + 5 ln ∣ x - 3 ∣ + 2 ln ∣ x ∣

= 3 ln ∣ x + 1 ∣ + 5 ln ∣ x - 3 ∣ + 2 ln ∣ x ∣

Add a constant to the solution

= 3 ln ∣ x + 1 ∣ + 5 ln ∣ x - 3 ∣ + 2 ln ∣ x ∣ + C

\int sin^{2} (\frac{nπ x}{l}) d x

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

\int \frac{2 x - 1}{x ^{2} - 7 x + 10} d x

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

What is the integral of (10x^2-8x-6)/(x^3-2x^2-3x) ?
The integral of (10x^2-8x-6)/(x^3-2x^2-3x) is 3ln|x+1|+5ln|x-3|+2ln|x|+C