Evaluate the following :
`int_(1)^(2)(x^(2))/(x^(2)+4x+3)dx`

Explanation

Let `I=int_(1)^(2)(x^(2))/(x^(2)+4x+3)dx=int_(1)^(2)((x^(2)+4x+3)-(4x+3))/(x^(2)+4x+3)dx`
`=int_(1)^(2)(1-(4x+3)/(x^(2)+4x+3))dx`
Let `(4x+3)/(x^(2)+4x+3)=(4x+3)/((x+3)(x+1))=(A)/(x+3)+(B)/(x+1)`
`therefore 4x+3=A(x+1)+B(x+3)`
Put `x+3=0`, i.e., `x=-3`, we get
`4(-3)+3=A(-2)+B(0)" "therefore A=(9)/(2)`
Put `x+1=0`, i.e., `x=-1`, we get,
`4(-1)+3=A(0)+B(2)" "therefore B=-(1)/(2)`
`therefore (4x+3)/(x^(2)+4x+3)=(((9)/(2)))/(x+3)+((-(1)/(2)))/(x+1)`
`therefore I=int_(1)^(2)[1-(((9)/(2)))/(x+3)+(((1)/(2)))/(x+1)]dx`
`=int_(1)^(2)1dx-(9)/(2)int_(1)^(2)(1)/(x+3)dx+(1)/(2)int_(1)^(2)(1)/(x+1)dx`
`=[x]_(1)^(2)-(9)/(2)[log|x+3|]_(1)^(2)+(1)/(2)[log|x+1|]_(1)^(2)`
`=(2-1)-(9)/(2)(log5-log4)+(1)/(2)(log3-log2)`
`=-1-(9)/(2)log((5)/(4))+(1)/(2)log((3)/(2)).`