इन समाकलनो का मान ज्ञात कीजिये
`int_(1)^(2) (5x^(2))/(x^(2)+4x+3)`

Explanation

यहाँ समाकलन `underset(1)overset(2)int (5x^(2))/(x^(2)+4x+3)` एक विषम भिन्न है इसलिए भाग कलां विधि से
`(5x^(2))/(x^(2)+4x+3)=5-((20x+15))/(x^(2)+4x+3)`
`=5-(5(4x+3))/((x+3)(x+1))`
`Rightarrow underset(1)overset(2)int (5x^(2))/(x^(2)+4x+3)dx=underset(1)overset(2)int [5-(5(4x+3))/((x+3)(x+1))]dx`
`Rightarrow underset(1)overset(2)int (5x^(2))/(x^(2)+4x+3)dx`
`=5underset(1)overset(2)int 1dx=5underset(1)overset(2) int (4x+3)/((x+3)(x+1))dx``=underset(1)overset(2)int (5x^(2))/(x^(2)+4x+3)dx=5[x]_(1)^(2)=5underset(1)overset(2)int ((4x+3))/((x+3)(x+1))dx`
`=underset(1)overset(2)int (5x^(2))/(x^(2)+4x+3)dx=5(2-1)-5underset(1)overset(2)int ((4x+3))/((x+3)(x+1))dx`
`Rightarrow underset(1)overset(2)int (5x^(2))/(x^(2)+4x+3)dx =5-5 underset(1)overset(2)int ((4+3))/((x+3)(x+1)).........(1)`
पुनः माना `(4x_3)/((x+3)(x+1))=(A)/(x+3)+B/(x+1)`
`Rightarrow 4x+3=A(x+1)+B(x+3)`
`x+1=0 rArr x=-1` रखने पर
`-1=2B rArr B=-1/2`
`x+3=0 rArr -3 ` रखने पर
`-9=-2A rArr =9/2`
`therefore (4x+3)/((x+3)(x+1))=(9)/(2(x+3))-1/(2(x+1))`
`rArr underset(1)overset(2)int (4x+3)/((x+3)(x+1))dx`
`=9/2underset(1)overset(2)int (1)/((x+3))dx-1/2 underset(1)overset(2)int (1)/((x+1))dx`
`=9/2[log(x+3)]_(1)^(2)-1/2[log(x+1)]_(1)^(2)`
`=9/2[log5-log4]-1/2[log3-log2]`
`=9/2 log""5/4-1/2""log ""3/2`
सेमि (1) से,
`underset(1)overset(2)int (5x^2)/(x^(2)+4x+3)dx=5-5 {9/2log""5/4-1/2log""3/2}=5-(45)/(2)log""5/4+5/2log""3/2`