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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:
Options:
of integration, then $f(x)$ is equal to:
Solution:
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Verified Answer
The correct answer is:
$\frac{1}{3}(x+4)$
Let $I=\int \frac{x+1}{\sqrt{2 x-1}} d x$
Put $\sqrt{2 x-1}=t$
$\begin{array}{l}
\therefore \quad 2 x-1=t^{2} \Rightarrow d x=t d t \\
I=\int \frac{\left(t^{2}+3\right)}{2} d t=\frac{t^{3}}{6}+\frac{3 t}{2}+C \\
=\frac{(2 x-1)^{\frac{3}{2}}}{6}+\frac{3}{2}(2 x-1)^{\frac{1}{2}}+C \\
=\sqrt{2 x-1}\left(\frac{x+4}{3}\right)+C \\
=f(x) \cdot \sqrt{2 x-1}+C
\end{array}$
Hence, $f(x)=\frac{x+4}{3}$
Put $\sqrt{2 x-1}=t$
$\begin{array}{l}
\therefore \quad 2 x-1=t^{2} \Rightarrow d x=t d t \\
I=\int \frac{\left(t^{2}+3\right)}{2} d t=\frac{t^{3}}{6}+\frac{3 t}{2}+C \\
=\frac{(2 x-1)^{\frac{3}{2}}}{6}+\frac{3}{2}(2 x-1)^{\frac{1}{2}}+C \\
=\sqrt{2 x-1}\left(\frac{x+4}{3}\right)+C \\
=f(x) \cdot \sqrt{2 x-1}+C
\end{array}$
Hence, $f(x)=\frac{x+4}{3}$
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