1.

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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