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

Explanation

Given,
`I= int_1^2 (5x^2)/(x^2+4x+3) dx`
`=>I = int _2^1(((6x^2)/x^+4x+3))dx`
`=>I=[I_1]_1^2`
Therefore, `I_1=int (5x^2)/ (x^2+4x+3)dx=>I_1= int 5+(-20x-15)/(x^2+4x+3)dx`
=`int 5dx-5 int(4x+3)/((x+3)(x+1))dx`
=`int 5x-5 int(4x+3)/((x+3)(x+1))dx`
`(4x+3)/((x+3)(x+1))=A/(x+3)+B/(x+1) =>(4x+3)/((x+3)(x+1))=(A(x+1)+B(x+3))/((x+1)(x+3))`
=>`4x+3=Ax+A+Bx+3B`
=>`4x+3=x(A+B)+A+3B`
By, Equating coefficient both sides, we get
`A+B=4.....(1)`
`A+3B=3....(2)`
Solve equation 1 and 2 we get,
`A=9/2,B=(-1/2)`
`I1=5x-5 int(A/(x+3)/B/(x+1))`
`I1=5x-5(9/2)log|x+3|-5(-1/2)log(x+1)`
`I1=5x-((45)/2)log|x+3|+(5/2)log(x+1)`
`I=[5x-((45)/2)log|x+3|+(5/2)log(x+1)]_1^2`
`I={5.2-(45)/2log|5|+(5/2)log(3)}-{5.1-(45)/2log|1|+(5/2)log(2)}`
`I=5-(45)/2log5+(5/2)log(3)+45log2+(5/2)log(2)`
`I=5-5/2(9log(5/4)-log(3/2))`