Method of Partial Fractions integral help

[King_Silver]

I have a question where f(x) = 20-2x^2/(x-1)(x+2)^2 and have solved for constants A,B and C.
A = 2
B = -4
C = -4
I have worked this out myself. Now I am told to compute the indefinite integral and I am getting this answer but apparently it is wrong and I don't understand how?
My answer: 2ln(abs(x-1))-4ln(x+2)-4ln(abs(x+2)^2)

Help?

 

[Euler2718]

Your constants are right but you're integration isn't correct.

$$\int -\frac{4}{(x+2)^{2}} \neq -4\ln|(x+2)^{2} |$$

 

[King_Silver]

Your constants are right but you're integration isn't correct.

$$\int -\frac{4}{(x+2)^{2}} \neq -4\ln|(x+2)^{2} |$$

Yea that is the part I am stuck on, I don't know what part of that integration is going wrong.

 

[Samy_A]

In general, for $p\neq-1$ and $a$ a constant, $\int (x+a)^pdx=\frac {1}{p+1} (x+a)^{p+1} + C$

 

[SammyS]

I have a question where f(x) = 20-2x^2/(x-1)(x+2)^2 and have solved for constants A,B and C.
A = 2
B = -4
C = -4
I have worked this out myself. Now I am told to compute the indefinite integral and I am getting this answer but apparently it is wrong and I don't understand how?
My answer: 2ln(abs(x-1))-4ln(x+2)-4ln(abs(x+2)^2)

Help?

Actually it's called the "Method of Partial Fractions".

To evaluate the integral, $\displaystyle \int -\,\frac{4}{(x+2)^{2}}dx \, ,\$ use the substitution u = x+2 .

 

[Ray Vickson]

I have a question where f(x) = 20-2x^2/(x-1)(x+2)^2 and have solved for constants A,B and C.
A = 2
B = -4
C = -4
I have worked this out myself. Now I am told to compute the indefinite integral and I am getting this answer but apparently it is wrong and I don't understand how?
My answer: 2ln(abs(x-1))-4ln(x+2)-4ln(abs(x+2)^2)

Help?

First: you have written
$$f(x) = 20 - \frac{2x^2}{(x-1)}(x+2)^2$$
Is that what you meant, or did you want
$$f(x) = \frac{20 - 2 x^2}{(x-1)(x+2)^2}?$$
If you meant the latter, you need to use parentheses, like this: (20 - 2 x^2)/[(x-1)(x+2)^2].

Second: what denominators go with the constants A,B,C? We can guess, but we should not need to.

Third: $\int (x+2)^{-2} \, dx$ does not involve logarithms.

 

[I like Serena]

Hi King_Silver! :)

What's the derivative of $\frac{1}{x+2}$?
Does that give us a clue?

 

[King_Silver]

Thanks everyone I actually realized my mistake it was fairly stupid :) fixed it now and got it right!
it was (4/(x+2))+2ln(abs(x-1))-4ln(abs(x+2))