Evaluate :`int_1^2 (5x^2)/(x^2+4x+3) dx`

Explanation

`"Let "I= int_(1)^(2)(5x^(2))/(x^(2)+4x+3)dx`
`" Here " (5x^(2))/(x^(2)+4x+3)=5-(20x +15)/(x^(2)+4x+3)`
`:. I=int_(1)^(2)(5-(20x+15)/(x^(2)+4x+3))dx`
`=int_(1)^(2)5dx-int_(1)^(2)(20x+15)/((x+1)(x+3))dx" "......(1)`
`" Let " (20x+15)/((x+1)(x+3))dx=(A)/((x+1))+(B)/((x+3))`
`rArr " "20x +15 =A(x+3)+B(x+1)`
x=-1 then -20 +15 =A (-1+3)+0
`rArr " " A=-(5)/(2)
x=-3 then -60 +15 =0+ B (-3+1)`
`rArr " " B=(45)/(2)`
`:. I = 5 int_(1)^(2) 1 dx +(5)/(2) int_(1)^(2) (1)/((x+1)) dx`
`-(45)/(2) int_(1)^(2) (1)/((x+3))dx`
[From equation (1)]`
`[5x +(5)/(2) log|x+1|-(45)/(2) log |x+3|]_(1)^(2)`
`=(10+(5)/(2)log|3|-(45)/(2)log |5|-5-(5)/(2)Log|2|+(45)/(2)log|4|)`

`I=5+(5)/(2) log |(3)/(2)|-(45)/(2)log |(5)/(4)|`