Question

\[\int \frac{(4v^2 + 4v + 1)dv}{v^3 + v^2}\] how?? o.O partial fractions??

Answer 1

assuming denom to be v^2 + v
its integral (4 + 1/(v^2 + v) dv)
shouldnt be much tough..denom is (v)(v+1)

Answer 2

\[\int\limits \frac{1+4 v+4 v^2}{v^3+v^2} \, dv\]
Split
\[\int\limits \frac{4 v^2+4 v+1}{v^2 (v+1)} \, dv\]
Partial fraction

Answer 3

More specifically,
\[\frac{4v^2+4v+1}{v^3+v^2}=\frac{4v^2+4v+1}{v^2(v+1)}=\frac{Av+B}{v^2}+\frac{C}{v+1}\]

Answer 4

i figured it out lol sorry

Answer 5

ohh am sorry i thought you by mistake wrote v^2+ v^2
never noticed it to be v^3..
hmmn
so bottom line is : you know any good eye specialist? :P