InterviewSolution
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InterviewSolution
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`int{log(logx)+(1)/((logx)^(2))}dx=x {f (x)-g(x)}+C`, thenA. `f(x)=log(logx),g (x)=(1)/(logx)`B. `f(x)=logx,g(x)=(1)/(logx)`C. `f(x)=(1)/(logx),(x)=log(logx)`D. `f(x)=(1)/(xlogx),g(x)=(1)/(logx)` |
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Answer» Correct Answer - a We have , `I=int{log(logx)+(1)/((logx)^(2))}dx` `rArr I= inte^(t)(logt+(1)/(t^(2)))dt`,where t = log x `rArrI=inte^(t)(logt+(1)/(t))dt+inte^(t)(-(1)/(t)+(1)/(t^(2)))dt` `rArrI=e^(t)logt +e^(t)(-(1)/(t))+C` `rArrI=x(log(logx)-(1)/(logx))+C` `:. f(x) =logx and g(x) = (1)/(logx)` |
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