Partial Fraction or Not

[markraz]

This problem, I solved not using partial fractions. I used u sub.
However I was curious if you could use partial fractions

\(\displaystyle \Large \frac{x^2+12}{x^2+4} \)

If it is solvable via Partial Fractions, I know I have to do long divsion as a preliminary step
due to M=N. Also I noticed the denominator is not factor-able.

I realize there are quicker ways to solve this, but just curious if something like this can be done via partial frac

thanks

 

[Deleted member 4993]

This problem, I solved not using partial fractions. I used u sub.
However I was curious if you could use partial fractions

\(\displaystyle \Large \frac{x^2+12}{x^2+4} \)

If it is solvable via Partial Fractions, I know I have to do long divsion as a preliminary step
due to M=N. Also I noticed the denominator is not factor-able.

I realize there are quicker ways to solve this, but just curious if something like this can be done via partial frac

thanks

It can be done - but you know that your answer will contain "i" - and then you'll have to transform those to real domain.

x² + 4 = (x + 2i)(x - 2i)

\(\displaystyle \Large \dfrac{x^2+12}{x^2+4} = 1 + \dfrac{2}{i}\left[\dfrac{1}{x-2i} - \dfrac{1}{x+2i}\right ]\)

Unnecessary hassle - I would say...

 

[markraz]

It can be done - but you know that your answer will contain "i" - and then you'll have to transform those to real domain.

x² + 4 = (x + 2i)(x - 2i)

\(\displaystyle \Large \dfrac{x^2+12}{x^2+4} = 1 + \dfrac{2}{i}\left[\dfrac{1}{x-2i} - \dfrac{1}{x+2i}\right ]\)

Unnecessary hassle - I would say...

thanks appreciate it