The value of `int_1^e((tan^(-1)x)/x+(logx)/(1+x^2))dxi s` `tane` (b) `tan^(-1)e` `tan^(-1)(1/e)` (d) none of these

Options

A. `tane`
B. `tan^(-1)e`
C. `tan^(-1)(1//e)`
D. none of these

Answer

B

Explanation

`int_(1)^(e)((tan^(-1)x)/x+(logx)/(1+x^(2)))dx`
`=int_(1)^(e)(tan^(-1)x)/x dx+int_(1)^(e)(logx)/(1+x^(2)) dx`
`=int_(1)^(e)(tan^(-1)x)/x dx+(logx tan^(-1)x)_(1)^(e)-int_(1)^(e)(tan^(-1)x)/x dx`
`=tan^(-1)e`