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

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

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

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

\frac{7}{3} ln ∣ x - 1 ∣ + \frac{5}{3} ln ∣ x + 2 ∣ - 3 ln ∣ x ∣ + C

Solution steps

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

Expand

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

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

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

= \frac{1}{3} ln ∣ x - 1 ∣ + \frac{2}{3} ln ∣ x + 2 ∣ - 3 ln ∣ x ∣ + 2 ln ∣ x - 1 ∣ + ln ∣ x + 2 ∣

Simplify

\frac{1}{3} ln ∣ x - 1 ∣ + \frac{2}{3} ln ∣ x + 2 ∣ - 3 ln ∣ x ∣ + 2 ln ∣ x - 1 ∣ + ln ∣ x + 2 ∣ : \frac{7}{3} ln ∣ x - 1 ∣ + \frac{5}{3} ln ∣ x + 2 ∣ - 3 ln ∣ x ∣

= \frac{7}{3} ln ∣ x - 1 ∣ + \frac{5}{3} ln ∣ x + 2 ∣ - 3 ln ∣ x ∣

Add a constant to the solution

= \frac{7}{3} ln ∣ x - 1 ∣ + \frac{5}{3} ln ∣ x + 2 ∣ - 3 ln ∣ x ∣ + C

f (x) = \frac{2}{x + 7}, (- 8, - 2)

\frac{d}{d x} (e^{x} arctan (x))

x (\frac{d y}{d x}) = \frac{1}{y ^{3}}

\frac{d y}{d x} = (x + y + 4)^{2}

\frac{e ^{x}}{5 + 5 x}

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