`int_(2)^(3)(x dx)/(x^(2)+1)`

Explanation

माना `I=int_(2)^(3)(x)/(x^(2)+1)dx`
माना `" "x^(2)+1=t`
`rArr" "2x dx = dt " "rArr" "xdx=(dt)/(2)`
`x = 2` तो t = 5 तथा x = 3 तो t = 10
`I=(1)/(2)int_(5)^(10)(1)/(t)dt=(1)/(2)[log|t|]_(5)^(10)`
`=(1)/(2)[log|10|-log|5|]`
`=(1)/(2)log|(10)/(5)|=(1)/(2)log2`