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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 theseA. `tane`B. `tan^(-1)e`C. `tan^(-1)(1//e)`D. none of these |
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Answer» Correct Answer - B `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` |
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