Question

how would you work this partial fraction problem? sqrt(2y^2-3)/y^2

Answer 1

partial decomp you mean?

Answer 2

\[\frac{\sqrt{2y^2-3}}{y^2}\] ???

Answer 3

your question is not complete.

Answer 4

yes that but with the dy as a variable and is a normal intergration

Answer 5

as in you put +C at the end

Answer 6

then this is the question
\[\int\frac{\sqrt{2y^2-3}}{y^2}dy\]

Answer 7

yes

Answer 8

\[\frac{\sqrt{2y^2-3}}{y^2}=\frac{1}{y^2}\sqrt{2y^2-3}\]
\[\frac{1}{y^2}\sqrt{2y^2-3}=\sqrt{\frac{1}{y^4}(2y^2-3)}\]

Answer 9

maybe as a start :)

Answer 10

thanks

Answer 11

but starts arent always good for me :)

http://www.wolframalpha.com/input/?i=int+sqrt%282y^2-3%29%2Fy^2+dy

looks to be a trig substitution

Answer 12

it tends to get messier from there ...

Answer 13

Try y = sin theta.....

Answer 14

Seems to work out quite nicely with a bit of jiggling.