`f(x)=int_1^x(tan^(-1)(t))/t dt \ AAx in R^+,` then find the value of `f(e^2)-f(1/(e^2))`

Answer

`pi`

Explanation

`f(x)=int_(1)^(x)(tan^(-1)(t))/t dt`
`:.f(1/x)=int_(1)^(1//x)(tan^(-1)(t))/tdt`
Put `=t=1//u`
`:.dt=-(du)/(u^(2))`
`:.f(1//x)=int_(1)^(x)("tan"^(-1)(1/u))/(1/u)(-1/(u^(2)))du`
`=-int_(1)^(x)("tan"^(-1)(1/u))/u du`
`=-int_(1)^(x)("cot"^(-1)(u))/u du`
`=-int_(1)^(x)(cot^(-1)(t))/t dt`
Now` f(x)-f(1//x)=int_(1)^(x)(tan^(-1)t+cot^(-1)t)/t dt`
`=int_(1)^(x) (pi)/2xx1/t dt`
`=(pi)/2 log (x)`
`:. f(e^(2))-f(1//e^(2))=(pi)/2log_(e)e^(2)=pi`