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`int(x^(2)(1-logx))/((logx)^(4)-x^(4))dx` equalsA. `(1)/(2)ln.(x)/(lnx)-(1)/(4)ln(ln^(2)x-x^(2))+C`B. `(1)/(4)ln((lnx-x)/(lnx+x))-(1)/(2)tan^(-1)((lnx)/(x))+C`C. `(1)/(4)ln((lnx-x)/(lnx+x))+(1)/(2)tan^(-1)((lnx)/(x))+C`D. `(1)/(4)(ln((lnx-x)/(lnx+x))+tan^(-1)((lnx)/(x)))+C` |
Answer» Correct Answer - B `I=int(x^(2)(1-logx))/((logx)^(4)-x^(4))dx` `=int(1-logx)/(x(((logx)/(x))^(4)-1))dx` Put `(logx)/(x)=t rArr (1-logx)/(x^(2))=dt` `therefore" "I=int(dt)/((t^(4)-1))=int(dt)/((t^(2)+1)(t^(2)-1))` `=(1)/(2)int((t^(2)+1)-(t^(2)-1))/((t^(2)+1)(t^(2)-1))dt` `=(1)/(2)(int(dt)/(t^(2)-1)-int(dt)/(t^(2)+1))=(1)/(2)((1)/(2)ln(t-1)/(t+1)-tan^(-1)t)` `=(1)/(4)ln((lnx-x)/(lnx+x))-(1)/(2)tan^(-1)((lnx)/(x))+C` |
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