Q99 / 150 · MHT CET 2022 (10 Aug Shift 2) · Mathematics

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

  • A
    $\frac{2}{3}(x+2)$
  • B
    $\frac{2}{3}(x-4)$
  • Correct answer
    C
    $\frac{1}{3}(x+4)$
  • D
    $\frac{1}{3}(x+1)$

Solution

$\begin{aligned} & \int \frac{x+1}{\sqrt{2 x-1}} \mathrm{~d} x=\int \frac{\frac{t+1}{2}+1}{\sqrt{t}} \mathrm{~d} t[\text { Let } 2 x-1=t] \\ & =\int \frac{t+3}{4+\sqrt{t}} \mathrm{~d} t=\frac{1}{4} \int \sqrt{t} \mathrm{~d} t+\frac{3}{4} \int \frac{\mathrm{d} t}{\sqrt{t}} \\ & =\frac{1}{4} \times \frac{2}{3} t^{3 / 2}+\frac{3}{4} \times 2 t^{1 / 2}+C=\frac{1}{6} \sqrt{t}\{t+9\}+C \\ & =\frac{1}{6} \sqrt{2 x-1}\{2 x-1+9\}+C \\ & =\frac{1}{3}(x+4) \sqrt{2 x-1}+C \\ & \Rightarrow f(x)=\frac{1}{3}(x+4)\end{aligned}$

Hence, the correct answer is option: 'C'.

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