What is the integral of (x^2+8)/(x^4-x^3-2x^2) ?

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

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integral of (x^2+8)/(x^4-x^3-2x^2)

\int \frac{x ^{2} + 8}{x ^{4} - x ^{3} - 2 x ^{2}} d x

2 ln ∣ x ∣ + \frac{4}{x} - 3 ln ∣ x + 1 ∣ + ln ∣ x - 2 ∣ + C

Solution steps

\int \frac{x ^{2} + 8}{x ^{4} - x ^{3} - 2 x ^{2}} d x

Take the partial fraction of

\frac{x ^{2} + 8}{x ^{4} - x ^{3} - 2 x ^{2}} : \frac{2}{x} - \frac{4}{x ^{2}} - \frac{3}{x + 1} + \frac{1}{x - 2}

= \int \frac{2}{x} - \frac{4}{x ^{2}} - \frac{3}{x + 1} + \frac{1}{x - 2} d 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{2}{x} d x - \int \frac{4}{x ^{2}} d x - \int \frac{3}{x + 1} d x + \int \frac{1}{x - 2} d x

\int \frac{2}{x} d x = 2 ln ∣ x ∣

\int \frac{4}{x ^{2}} d x = - \frac{4}{x}

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

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

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

Simplify

= 2 ln ∣ x ∣ + \frac{4}{x} - 3 ln ∣ x + 1 ∣ + ln ∣ x - 2 ∣

Add a constant to the solution

= 2 ln ∣ x ∣ + \frac{4}{x} - 3 ln ∣ x + 1 ∣ + ln ∣ x - 2 ∣ + C

x \to + x lim (\frac{x ^{2} - 1}{x - 1})

d er i v a t i v e f (x) = \frac{7}{( x ^{2} - 10 ) ^{4}}

d er i v a t i v e f (x) = \frac{4}{x} - 16 sec (x)

\int - 3 x^{4} - 6 x^{2} + 8 d x

x \to 0 lim (x csc (π x))

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