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`int(dx)/(x(logx)^(m))` is equal to thenA. `((logx)^(m))/(m)+C`B. `((logx)^(m-1))/(m-1)+C`C. `((logx)^(1-m))/(1-m)+C`D. `((logx)^(1-m))/(m)+C` |
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Answer» Correct Answer - C `int(1)/(x(logx)^(m))dx` Let `log x-t rArr (1)/(x)=(dt)/(dx) rArr dx=xdx` `therefore" "int(1)/(x(logx)^(m))dx=int(1)/(x(t)^(m))xdt int t^(-m)dt` `=(t^(-m+1))/(-m+1)+C=((logx)^(1-m))/(1-m)+C` |
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