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The value of `intxlogx(logx-1)dx` is equal toA. `2(xlogx-x)^(2)+C`B. `(1)/(2)(xlogx-x)^(2)+C`C. `(x logx)^(2)+C`D. `(1)/(2)(x logx)^(3)+C` |
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Answer» Correct Answer - b Let `I=intx logx (logx-1)dx=intlogx(xlogx-x)dx` `rArrI=int(xlogx-x)d (xlogx-x)=((xlogx-x)^(2))/(2)+C` |
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