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

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

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

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

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

Solution steps

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

Expand

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

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

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

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

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

Simplify

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

Add a constant to the solution

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

- \frac{x}{6 y}

\frac{d ^{2}}{d x ^{2}} ((x^{2} + 1)^{5})

x y^{^{'}} + 2 y = - 4 cos (x)

\int arcsin (\frac{2 x}{5}) d x

\int \frac{1}{1 - x ^{4}} d x

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