Indefinite integral

Indefinite integral

I need to calculate $$\int\frac{dx}{(x^{4}-16)^{2}}$$. I made the integral to sum of this two: $$\int\frac{dx}{x^{4}-16}$$ - $$\int\frac{x^{4}-17}{(x^{4}-16)^{2}}dx$$.
The first one is prety simple but i dont have any idea how to solve the second one. Any suggestions would be helpful!
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Re: Indefinite integral

You need "partial fractions". $$x^4- 16= (x^2- 4)(x^2+ 4)= (x- 2)(x+ 2)(x^2+ 4)$$.
$$(x^2- 16)^2= (x- 2)^2(x+ 2)^2(x^2+ 4)^2$$.

That can be expanded in partial fractions as
$$\frac{A}{x- 2}+ \frac{B}{(x- 2)^2}+ \frac{C}{x+ 2}+ \frac{D}{(x+ 2)^2}+ \frac{Ex+ F}{x^2+ 4}+ \frac{Gx+ H}{(x^2+ 4}$$.

You will need to solve for A, B, C, D, E, F, G, and H.

With $$\frac{1}{(x^4- 16)^2}= \frac{A}{x- 2}+ \frac{B}{(x- 2)^2}+ \frac{C}{x+ 2}+ \frac{D}{(x+ 2)^2}+ \frac{Ex+ F}{x^2+ 4}+ \frac{Gx+ H}{(x^2+ 4}$$ multiply on both sides by $$(x^4- 16)^2$$. On the left we have, of course, 1. On the right one or two of the various factors will cancel leaving
$$1= A(x-2)(x+ 2)^2(x^2+ 4)^2+ B(x+ 2)^2(x^2+ 4)^2+ C(x- 2)^2(x+ 2)(x^2+ 4)^2+ D(x- 2)^2(x^2+ 4)^2+ (Ex+ F)(x- 2)^2(x+ 2)^2(x^2+ 4)+ (Gx+ H)(x- 2)^2(x+ 2)^2$$.

That equation is to be true for all x. Taking 8 values for x will give us 8 equations to solve for A, B, C, D, E, F, G, and H. In particular, if we take x= 2, since every coefficient except that of B has a factor of x- 2, we have simply $$1= B(4^2)(8^2)= 1024B$$ so $$B= \frac{1}{1024}$$. Similarly, taking x= -2, we have simply
$$1= D(-4)^2(8^2)= 1024D$$ so D is also $$\frac{1}{1024}$$.

The others can' t be reduced so simply but replacing x with 6 different numbers (other than 2 or -2) gives 6 equations to solve for A, C, E, F, G, and H.

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