What is $\int \frac{2}{(y^2+1)(y-1)^2}\,\mathrm{d}y$?

Question

I tried to do the partial fraction method but couldn't do it. Maybe I made a mistake or something. i tried [(Ay+B)(y-1)^2] + [C(y^2 +1)(y-1)] + [D(y^2 +1)] But i couldnt find A,b,c and d.

Answer 1

Use that $$\frac{2}{(y^2+1)(y-1)^2}=\frac{y}{y^2+1}+\frac{1}{(y-1)^2}-\frac{1}{y-1}$$

Answer 2

For problem like this you need to do partial fractions method.

We have $$\frac {2}{(y^2+1)(y-1)^2} = \frac {Ay+B}{y^2+1} +\frac {Cy+D}{(y-1)^2}$$

Once you have found your $A, B, C, D$ the integral are easy to solve.

Answer 3

In order to simplify the equation we have: $$2/((y^2+1)〖(y-1)〗^2 )=(ay+b)/(y^2+1)+(cy+d)/〖(y-1)〗^2 .$$ Then we have $$2=(ay+b) (y-1)^2+(cy+d) 〖(y〗^2+1)=(a+c) y^3+(b-2a+d) y^2-(2b-a-c)y+(b+d).$$ So, $$a+c=0, b+d-2a+0, 2b-a-c=0$$, and $b+d=2$.

We have $2b=0$, then $d=2$, then $a=1$, then $c= -1$. So $$2/((y^2+1)〖(y-1)〗^2 )=y/(y^2+1)+(-y+2)/〖(y-1)〗^2 =1/2×2y/(y^2+1)-1/(y-1)+1/〖(y-1)〗^2$$ . In this way the answer is $$1/2 ln⁡(y^2+1)-ln⁡(y-1)-1/(y-1).$$