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If `int (x+1)/(sqrt(2x-1))dx = f(x)sqrt(2x-1)+C`, where C is a constant of integration, then f(x) is equal toA. `(2)/(3)(x+2)`B. `(1)/(3)(x+4)`C. `(2)/(3)(x-4)`D. `(1)/(3)(x+1)` |
Answer» Correct Answer - B ....(i) We have, `int(x+1)/(sqrt(2x-1))dx = f(x)sqrt(2x-1)+C` Let `I=int(x+1)/(sqrt(2x-1))dx` Put `2x - 1 = t^(2) rArr 2dx = 2tdt rArr dx = tdt` `I=int((t^(2)+1)/(2)+1)/(t)tdt = (1)/(2)int(t^(2)+3)dt" "[therefore 2x-1=t^(2) rArr x = (t^(2)+1)/(2)]` `=(1)/(2)((t^(3))/(3)+3t)+C=(t)/(6)(t^(2)+9)+C` `=(sqrt(2x-1))/(6)(2x-1+9)+C" "[therefore t = sqrt(2x-1)]` `=(sqrt(2x-1))/(6)(2x+8)+C` `=(x+4)/(3)sqrt(2x-1)+C` On comparing it with Eq. (i), we get `f(x)=(x+4)/(3)` |
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