Tricky Partial Fractions Question

[ardentmed]

Hey guys,

Here is another pair of questions that I'm doubting at the moment:
View attachment 2798

I used partial fractions for A and got (Bx+C)/x^2 + Ax/(x-1)^2 + Dx(x-1) which led me to compute A=1, B=0, C= -1, and D=0, which already sounds off. Do you guys have any suggestions?

Also, for 5b, I calculated B= -1, C=-1, A=2, and a final answer of 2ln(x) - (1/2)ln(x^2 + 3) - (1/3) tan^-1(x/√3) + C. Any tips for this one?

Thanks in advance. I really appreciate the help.

 

[Prove It]

Hey guys,

Here is another pair of questions that I'm doubting at the moment:
View attachment 2798

I used partial fractions for A and got (Bx+C)/x^2 + Ax/(x-1)^2 + Dx(x-1) which led me to compute A=1, B=0, C= -1, and D=0, which already sounds off. Do you guys have any suggestions?

Also, for 5b, I calculated B= -1, C=-1, A=2, and a final answer of 2ln(x) - (1/2)ln(x^2 + 3) - (1/3) tan^-1(x/√3) + C. Any tips for this one?

Thanks in advance. I really appreciate the help.

For the first I would use as my partial fraction decomposition:

$\displaystyle \begin{align*} \frac{1}{ x^2 \left( x - 1 \right) ^2} \equiv \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x -1 } + \frac{D}{ \left( x - 1 \right) ^2} \end{align*}$

For the second

$\displaystyle \begin{align*} \frac{x^2 - x + 6}{x \left( x^2 + 3 \right) } &\equiv \frac{A}{x} + \frac{B\,x + C}{x^2 + 3} \end{align*}$