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Question:

If $\int \frac{x+1}{\sqrt{2 x-1}} \mathrm{~d} x=f(x) \sqrt{2 x-1}+\mathrm{C}$, where $\mathrm{C}$ is a constant

of integration, then $f(x)$ is equal to:

  1. (1) $\frac{1}{3}(x+1)$

  2. (2) $\frac{2}{3}(x+2)$

  3. (3) $\frac{2}{3}(x-4)$

  4. (4) $\frac{1}{3}(x+4)$


Correct Option: , 3

Solution:

Let $I=\int \frac{x+1}{\sqrt{2 x-1}} d x$

Put $\sqrt{2 x-1}=t$

$\therefore \quad 2 x-1=t^{2} \Rightarrow d x=t d t$

$I=\int \frac{\left(t^{2}+3\right)}{2} d t=\frac{t^{3}}{6}+\frac{3 t}{2}+C$

$=\frac{(2 x-1)^{\frac{3}{2}}}{6}+\frac{3}{2}(2 x-1)^{\frac{1}{2}}+C$

$=\sqrt{2 x-1}\left(\frac{x+4}{3}\right)+C$

$=f(x) \cdot \sqrt{2 x-1}+C$

Hence, $f(x)=\frac{x+4}{3}$

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