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

Explanation

यहाँ, अंश और हर की घात 4 है। इसलिए हम `x^(2)=t` रखकर भाग करते हैं।
`((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)`
अब ,`" " int((x^(2)+1)(x^(2)+2))/((x^(2)+3)(x^(2)+4))dx`
`=int1dx-int((4t+10))/(t^(2)+7t+12)dx`
`=int1dx-int(4t+10)/((t+3)(t+4))dx" ...(1)"`
माना `(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,"तो ",-12+10=0+B(1),rArr.,B=-2),(t=-4,"तो ",-16+10=A(-1)+0,rArr.,A=6):}`
`therefore" "(4t+10)/((t+4)(t+3))=(6)/(t+4)-(2)/(t+3)`
यह मान समीकरण (1 ) में रखने पर,
`int((x^(2)+1)(x^(2)+2))/((x^(2)+3)(x^(2)+4))dx`
`=int1dx-int((6)/(t+4)-(2)/(t+3))dx`
`=int1dx-int[(6)/(x^(2)+4)-(2)/((x^(2)+3))]dx`
(`t=x^(2)` रखने पर )
`rArr" "int((x^(2)+1)(x^(2)+2))/((x^(2)+3)(x^(2)+4))dx`
`=int1dx-int[(6)/(x^(2)+2^(2))-(2)/(x^(2)+(sqrt3)^(2))]dx`
`=x-6((1)/(2)tan^(-1).(x)/(2))+2((1)/(sqrt3)tan^(-1).(x)/(sqrt3))+C`
`=x-3tan^(-1).(x)/(2)+(2)/(sqrt3)tan^(-1).(x)/(sqr3)+C`