Evaluate : `int(x^2)/((x^2+4)(x^2+9))dx`

Explanation

`Let (x^(2))/((x^(2)+4)(x^(2)+9))=(y)/((y+4)(y+9)),"where "x^(2)=y.`
`Let (y)/((y+4)(y+9))=(A)/((y+4))+(B)/((y+9)).`
then`,y-=A(y+9)+B(y+4).`
Putting `y=-4`on both sides of(i) , we get `A=(-4)/(5).`
Putting `y=-9`on both sides of (i) , we get `B=(9)/(5).`
`therefore (y)/((y+4)(y+9))=(-4)/(5(y+4))+(9)/(5(y+9))`
`implies(x^(2))/((x^(2)+4)(x^(2)+9))=(-4)/(5(x^(2)+4))+(9)/(5(x^(2)+9))`
`implies int(x^(2))/((x^(2)+4)(x^(2)+9))dx=(-4)/(5)int(dx)/((x^(2)+4))+(9)/(5(x^(2)+9))=((-4)/(5)xx(1)/(2))tan^(-1)""(x)/(2)+((9)/(5)xx(1)/(3))tan^(-1)""(x)/(3)=(-2)/(5)tan^(-1)++(x)/(2)+(3)/(5)tan^(-1)""(x)/(3)+C.`