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The value of `int_(0)^(oo)(logx)/(a^(2)+x^(2))dx` isA. `(2piloga)/(a)`B. `(pi log a)/(2a)`C. `pi loga`D. 0 |
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Answer» Correct Answer - B `I=int_(0)^(oo)(logx)/(a^(2)+x^(2))dx` Put `x=(a^(2))/(y),dx=-(a^(2))/(y^(2))dy` `therefore" "I=int_(oo)^(0)(log((a^(2))/(y)))/(a^(2)+(a^(4))/(y^(2)))((-a^(2))/(y^(2)))dy` `" "=int_(oo)^(0)((loga^(2)-logy))/(a^(2)+y^(2))(-dy)` `" "=log(a^(2))int_(0)^(oo)(dy)/(a^(2)+y^(2))-int_(0)^(oo)(logy)/(a^(2)+y^(2))dy` `" "log(a^(2))(1)/(a)(tan^(-1)((y)/(a)))_(0)^(oo)-I` `rArr" "2I=(2loga)/(a).(pi)/(2)` `rArr" "I=(piloga)/(2a)` |
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