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`int_- 1^1(e^(-1/ x))/(x^2(1+e^(-2/ x)))dx` is equal to :A. `(pi)/2=2tan^(-1)e`B. `(pi)/2-2cot^(-1)e`C. `2tan^(-1)e`D. `pi-2tan^(-1)e` |
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Answer» Correct Answer - D `I=int_(-1)^(1)(e^(-1//x))/(x^(2)(1+e^(-2//x)))dx=2 int_(0)^(1)(e^(-1//x))/(x^(2)(1+e^(-2//x)))dx` Put `e^(-1/x)=t` `implies 1/(x^(2))e^(-1/x)dx=dt` `:. I=2 int_(0)^(1//2) (dt)/(1+t^(2))` `=2[tan^(-1)t]_(0)^(1//e)` `=2"tan"^(-1)1/e` `=2[(pi)/2-tan^(-1)e]` `=pi-2 tan^(-1) e` |
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