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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

\int 1 d x = x

\int \frac{3}{2 ( x + 2 )} d x = \frac{3}{2} ln ∣ x + 2 ∣

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

\int \frac{3}{2 ( x - 2 )} d x = \frac{3}{2} ln ∣ x - 2 ∣

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

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

Add a constant to the solution

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

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