1.

` int _(0)^(3) (3x+1)/(x^(2)+9) dx = `A. `log(2sqrt(2))+pi/12`B. `log(2sqrt(2))+pi/2`C. ` log (2sqrt(2)) +pi/6`D. ` log (2sqrt(2))+pi/3`

Answer» Correct Answer - A
`int_(0)^(3)(3x+1)/(x^(2)+9)dx=(3)/(2)int_(0)^(3)(2x)/(x^(2)+9)dx+int_(0)^(3)(dx)/(x^(2)+9)`
`=[(3)/(2)log(x^(2)+9)+(1)/(3)tan^(-1)((x)/(3))]_(0)^(3)`
`=(3)/(2)(log18-log9)+(1)/(3)((pi)/(4))`
`=(3)/(2)log2+(pi)/(12)=log(2sqrt2)+(pi)/(12)`


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