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

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

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integral of (2x^3-4x-8)/(x(x-1)(x^2+4))

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

- 2 ln ∣ x - 1 ∣ + 2 ln ∣ x ∣ + ln x^{2} + 4 + 2 arctan (\frac{x}{2}) + C

Solution steps

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

Apply u-substitution

: \int \frac{- 4 u + 2 ( u + 1 ) ^{3} - 12}{u ( u + 1 ) ( u ^{2} + 2 u + 5 )} d u

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

Take the partial fraction of

\frac{- 4 u + 2 ( u + 1 ) ^{3} - 12}{u ( u + 1 ) ( u ^{2} + 2 u + 5 )} : - \frac{2}{u} + \frac{2}{u + 1} + \frac{2 u + 6}{u ^{2} + 2 u + 5}

= \int - \frac{2}{u} + \frac{2}{u + 1} + \frac{2 u + 6}{u ^{2} + 2 u + 5} d u

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}{u} d u + \int \frac{2}{u + 1} d u + \int \frac{2 u + 6}{u ^{2} + 2 u + 5} d u

\int \frac{2}{u} d u = 2 ln ∣ u ∣

\int \frac{2}{u + 1} d u = 2 ln ∣ u + 1 ∣

\int \frac{2 u + 6}{u ^{2} + 2 u + 5} d u = ln u^{2} + 2 u + 5 + 2 arctan (\frac{1}{2} (u + 1))

= - 2 ln ∣ u ∣ + 2 ln ∣ u + 1 ∣ + ln u^{2} + 2 u + 5 + 2 arctan (\frac{1}{2} (u + 1))

Substitute back

u = x - 1

= - 2 ln ∣ x - 1 ∣ + 2 ln ∣ x - 1 + 1 ∣ + ln (x - 1)^{2} + 2 (x - 1) + 5 + 2 arctan (\frac{1}{2} (x - 1 + 1))

Simplify

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

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

Add a constant to the solution

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

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