How is it possible that $\int\frac{dy}{(1+y^2)(2+y)}$ = $\frac{1}{5}\int\frac{dy}{y+2}-\int\frac{ydy}{1+y^2}+\int\frac{2dy}{1+y^2} $?

Question

Suppose we have a fraction

$$I=\int\frac{dy}{(1+y^2)(2+y)}$$

How is it possible that

$$5I = \int\frac{dy}{y+2}-\int\frac{ydy}{1+y^2}+\int\frac{2dy}{1+y^2} ?$$

How are they using partial fractions to do this?

Answer 1

Setting $$\frac{1}{(1+y^2)(2+y)}=\frac{A}{y+2}+\frac{By+C}{1+y^2}$$ gives $$1=A(1+y^2)+(y+2)(By+C),$$ i.e. $$0y^2+0y+1=(A+B)y^2+(C+2B)y+A+2C$$ Then, solve the following system : $$0=A+B,\quad 0=C+2B,\quad 1=A+2C$$

Answer 2

$$\begin{align*} \frac1{y+2}-\frac{y}{1+y^2}+\frac2{1+y^2}&=\frac1{y+2}+\frac{2-y}{1+y^2}\\ &=\frac{(1+y^2)+(4-y^2)}{(y+2)(1+y^2)}\\ &=\frac5{(y+2)(1+y^2)} \end{align*}$$