MHB How to Solve this Partial Fraction Decomposition Problem?

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The discussion revolves around solving the integral $$\int \frac{7dx}{x(x^2+8)^2}$$ using partial fraction decomposition. The initial setup proposed includes terms for A, B, C, D, and E, with A, B, and C values partially calculated. A participant points out that while A and B are correct, the value for C is incorrect. To find the remaining unknowns, it is suggested to substitute different values for x to create simultaneous equations. This method will help in determining the correct values for C, D, and E.
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$$\int \frac{7dx}{x(x^2+8)^2}$$

so I am thinking its going to be set up like: $$\frac{A}{x} + \frac{Bx + C}{x^2 + 8} + \frac{Dx + E}{(x^2 + 8)^2}$$
Practice problem I'm stuck on.
so I cleared fractions and got A = 7/64 , b = -7/64 and C = 105/64 and now I'm lost... can anyone work this problem for me so I can see what's going on? Thanks in advance.
 
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shamieh said:
$$\int \frac{7dx}{x(x^2+8)^2}$$

so I am thinking its going to be set up like: $$\frac{A}{x} + \frac{Bx + C}{x^2 + 8} + \frac{Dx + E}{(x^2 + 8)^2}$$
Practice problem I'm stuck on.
so I cleared fractions and got A = 7/64 , b = -7/64 and C = 105/64 and now I'm lost... can anyone work this problem for me so I can see what's going on? Thanks in advance.

Hi shamieh, :)

I think your value for C is not correct. However the other two values are correct.

\[\frac{7dx}{x(x^2+8)^2}=\frac{A}{x} + \frac{Bx + C}{x^2 + 8} + \frac{Dx + E}{(x^2 + 8)^2}\]

\[\Rightarrow A(x^2+8)^2+(Bx+C)(x^2+8)x+(Dx+E)x=7\]

Substituting different values for $x$ you can get three simultaneous equations (assuming you already found the values A and B correctly) which can be used to find the remaining unknowns, $C,\,D\mbox{ and }E$.
 
There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...

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