`int((x^2+1)(x^2+2))/((x^2+3)(x^2+4))dx`

Explanation

Hence the degree of numerator and denominator both are 4. So, we put `x^(2)`=t and divide.
`((x^(2)-1)(x^(2)-2))/((x^(2)+3)(x^(2)+4))=((t+1)(t+2))/((t+3)(t+4))`
` =(t^(2)+3t+2)/(t^(2)+7t+12)=1-(4t+10)/(t^(2)+7t+12).`
`" Now " int((x^(2)+1)(x^(2)+2))/((x^(2)+3)(x^(2)+4))dx`
`= int1dx -int((4t+10))/(t^(2)+7t+12)dx`
` = int 1dx-int (4t+10)/((t+3)(t+4))dx......(1)`
`" Let " (4t+10)/((t+4)(t+3)) =(A)/((t+4)) +(B)/((t+3))`
` rArr (4t+10)/((t+3)(t+4))=(A(t+3)+B(t+4))/((t+4)(t+3))`
`rArr 4t+10 =A(t+3) +B(t+4)`
t=-3 then -12 +10 =0 + A(-1) +0 `rArr` b=-2
t=-4 then - 16+10=A(-1) +0 `rArr ` A=6
`:. (4t+10)/((t+4)(t+3)) =(6)/(t+4)-(2)/(t+3)`
Put these values in equation (1).
`int ((x^(2)+1)(x^(2)+2))/((x^(2)+3)(x^(2)+4)) dx`
` = int 1dx - int ((6)/(t+4)-(2)/(t+3))dx`
` = int 1dx - int [(6)/(x^(2)+4)-(2)/((x^(2)+3))]dx`
`" (put t= "x^(2))`
`rArr int((x^(2)+1)(x^(2)+2))/((x^(2)+3)(x^(2)+4))dx`
`=int 1dx - int[(6)/(x^(2)-2^(2))-(2)/(x^(2)+(sqrt(3))^(2))]dx`
` =x-6 ((1)/(2)tan^(-1). (x)/(2)) +2 ((1)/(sqrt(3))tan^(-1) .(x)/(sqrt(3)))+c`
` =x-3 tan^(-1) .(x)/(2) +(2)/(sqrt(3)) tan^(-1) .(x)/(sqrt(3))+c`