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`((x+1)(x+logx)^(2))/(x)` |
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Answer» `int((x+1)(x+logx)^(2))/(x)dx" माना "x+logx=t` `=intt^(2)dt=(1)/(3)t^(3)+c" "rArr (1+(1)/(x))dx=dt` `=(1)/(3)(x+logx)^(3)+c" "rArr ((x+1)/(x))dx=dt` |
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