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

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Aicharya · Accepted Answer

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

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,…

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