Question

Integrate the following by Integration by partial fraction
(y^2+4)/[y(3y^2+4y-4)

Answer 1

\[\frac{A}{y}+\frac{By+c}{3y^2+4y-4}=\frac{(y^2+4)}{y(3y^2+4y-4)}\]\[\implies A(3y^2+4y-4)+(By+c)(y)=y^2+4\]\[\implies (3A+B) y^2+(4A+c)y-4A=y^2+4\]comparing we get
3A+B=1
4A+c=0
-4A=4
=>A=-1,c=4,B=4

Answer 2

so ur equation reduces to \[\int\limits \left( \frac{-1}{y}+\frac{4y+4}{3y^2+4y-4} \right)dy\]

Answer 3

solve it further & tell the result

Answer 4

Hint: The first (of three) integrals is\[-\int\limits_{}^{}\frac{ dy }{ y }\]

Can you evaluate this?

Answer 5

yes...
\[-\ln \left| y \right|+C\]

Answer 6

result
\[1/2\ln \left| y+2 \right|+5/6\ln \left| 3y-2 \right|-\ln \left| y \right|+C\]

Answer 7

I am no sure about but thanks lot guys for great help....