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

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

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

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

\frac{1}{2} ln \frac{x}{2} + 1 + \frac{1}{2 ( x + 2 )} + \frac{1}{2} ln \frac{x}{2} - 1 - \frac{1}{2 ( x - 2 )} + C

Solution steps

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

Apply Integral Substitution

= \int \frac{u ^{3}}{( u ^{2} - 1 ) ^{2}} d u

Take the partial fraction of

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

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

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

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

\int \frac{1}{2 ( u - 1 )} d u = \frac{1}{2} ln ∣ u - 1 ∣

\int \frac{1}{4 ( u - 1 ) ^{2}} d u = - \frac{1}{4 ( u - 1 )}

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

Substitute back

u = \frac{x}{2}

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

Simplify

\frac{1}{2} ln \frac{x}{2} + 1 - (- \frac{1}{4 ( \frac{x}{2} + 1 )}) + \frac{1}{2} ln \frac{x}{2} - 1 - \frac{1}{4 ( \frac{x}{2} - 1 )} : \frac{1}{2} ln \frac{x}{2} + 1 + \frac{1}{2 ( x + 2 )} + \frac{1}{2} ln \frac{x}{2} - 1 - \frac{1}{2 ( x - 2 )}

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

Add a constant to the solution

= \frac{1}{2} ln \frac{x}{2} + 1 + \frac{1}{2 ( x + 2 )} + \frac{1}{2} ln \frac{x}{2} - 1 - \frac{1}{2 ( x - 2 )} + C

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