`T h ev a l u eofint_0^1(2x^2+3x+3)/((x+1)(x^2+2x+2))dxi s` `pi/4+2log2-tan^(-1)2` `pi/4+2log2-tan^(-1)1/3` `2log2-cot^(-1)3` (d) `-pi/4+log4+cot^(-1)2`

Options

A. `(pi)/4+2log2-tan^(-1)2`
B. `(pi)/4+2log 2-"tan"^(-1)1/3`
C. `2 log 2 -cot^(-1)3`
D. `-(pi)/4+log4+cot^(-1)2`

Answer

A, C, D

Explanation

`I=int_(0)^(1)(2x^(2)+3x+3)/((x+1)(x^(2)+2x+2))dx`
`=int_(0)^(1)(2(x^(2)+2x+2)-(x+1))/((x+1)(x^(2)+2x+2))dx`
`=int_(0)^(1)(2/(x+1)-1/(x^(2)+2x+2))dx`
`=[2log(x+1)-tan^(-1)(x+1)]_(0)^(1)` ……………1
`=2log2-tan^(-1)2+tan^(-1)1`
`=2log2-tan^(-1)2+(pi)/4`
`=log4-((pi)/2-cot^(-1)2)+(pi)/4`
`=-(pi)/4+log4+cot^(-1)2`
From equation 1
`I=2log2-tan^(-1)((2-1)/(1+2xx1))`
`=2 log 2-"tan"^(-1)1/3`
`=2 log 2-cot^(-1)3`