What is the integral of (4x^5-4)/(x^2-4) ?

The integral of (4x^5-4)/(x^2-4) is x^4+8x^2+33ln|x+2|+31ln|x-2|+C

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integral of (4x^5-4)/(x^2-4)

\int \frac{4 x ^{5} - 4}{x ^{2} - 4} d x

x^{4} + 8 x^{2} + 33 ln ∣ x + 2 ∣ + 31 ln ∣ x - 2 ∣ + C

Solution steps

\int \frac{4 x ^{5} - 4}{x ^{2} - 4} d x

Long division

\frac{4 x ^{5} - 4}{x ^{2} - 4} : 4 x^{3} + 16 x + \frac{64 x - 4}{x ^{2} - 4}

= \int 4 x^{3} + 16 x + \frac{64 x - 4}{x ^{2} - 4} 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 4 x^{3} d x + \int 16 x d x + \int \frac{64 x - 4}{x ^{2} - 4} d x

\int 4 x^{3} d x = x^{4}

\int 16 x d x = 8 x^{2}

\int \frac{64 x - 4}{x ^{2} - 4} d x = 33 ln ∣ x + 2 ∣ + 31 ln ∣ x - 2 ∣

= x^{4} + 8 x^{2} + 33 ln ∣ x + 2 ∣ + 31 ln ∣ x - 2 ∣

Add a constant to the solution

= x^{4} + 8 x^{2} + 33 ln ∣ x + 2 ∣ + 31 ln ∣ x - 2 ∣ + C

n = 1 \sum \infty \frac{n}{1 + n ^{2}}

\frac{\partial}{\partial y} (ln (e^{x} + e^{y}))

y^{^{''}} + 8 y^{^{'}} + 41 y = 0

arccos (1 - x^{2})

\frac{d}{d x} (\frac{( x - 1 )}{2})

What is the integral of (4x^5-4)/(x^2-4) ?
The integral of (4x^5-4)/(x^2-4) is x^4+8x^2+33ln|x+2|+31ln|x-2|+C