Concept:

\(\rm \int\frac{1}{x^2+a^2}dx=\frac{1}{a}tan^{-1}\frac{x}{a}+C\)

Calculation:​

\(\rm \int\frac{y^2}{(y^2+2)(y^2+3)}dy\)

Substitute y² = x in the above integrand

⇒ \(\rm \frac{y^2}{(y^2+2)(y^2+3)}=\frac{x}{(x+2)(x+3)}\)

The integrand is a proper rational fraction. So, by using the form of partial fraction, we write

⇒\(\rm \frac{x}{(x+2)(x+3)}=\frac{A}{x+2}+\frac{B}{x+3}\)

⇒ x = (A + B)x+ 3A +2B

By comparing coefficient of x and constant terms on both sides, we get A + B = 1 and  3A + 2B = 0

By solving these equation, we get  A = -2  and  B = 3

⇒\(\rm \frac{x}{(x+2)(x+3)}=\frac{-2}{x+2}+\frac{3}{x+3}\) 

Now put x = y² in the above equation

⇒\(\rm \frac{y^2}{(y^2+2)(y^2+3)}=\frac{-2}{y^2+2}+\frac{3}{y^2+3}\)

⇒ \(\rm \int \frac{y^2}{(y^2+2)(y^2+3)}dy=\int\frac{-2}{y^2+2}dy+\int\frac{3}{y^2+3}dy\)

⇒ \(\rm \int \frac{y^2}{(y^2+2)(y^2+3)}dy= \frac{-2}{\sqrt{2}}tan^{-1}\frac{y}{\sqrt{2}}+\frac{3}{\sqrt{3}}tan^{-1}\frac{y}{\sqrt{3}}+C\)   

⇒ \(\rm \int \frac{y^2}{(y^2+2)(y^2+3)}dy= {-\sqrt{2}}tan^{-1}\frac{y}{\sqrt{2}}+{\sqrt{3}}tan^{-1}\frac{y}{\sqrt{3}}+C\)