If `alpha=int_0^1(e^(9x+3tan^-1x))((12+9x^2)/(1+x^2))dx`
where `tan^-1x` takes only principal value, then the value of `(log_eabs(1+alpha)-(3pi)/4)` is

Answer

9

Explanation

`alpha=int_(0)^(1)e^(9x+3tan^(-1)x)((12+9x^(2))/(1+x^(2)))dx`
Let `9x+3tan^(-1)x=t`
`implies (9-3)/(1+x^(2))dx=dt`
`implies((12+9x^(2))/(1+x^(2)))dx=dt`
`impliesalpha=int_(0)^(9+(3pi)/4)e^(t)dt=e^(9+(3pi)/4)-1`
`implies log_(e)(1+alpha)=9=(3pi)/4`