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JEE Mains · Maths · STD 12 - 7.1 indefinite integral

If \(\int {\frac{{x + 1}}{{\sqrt {2x - 1} }}} dx = f\left( x \right)\,\sqrt {2x - 1}  + C\) , where \(C\) is a constant of integration of integration, then \(f(x)\) is equal to

  1. A \(\frac{1}{3}\,\left( {x + 1} \right)\)
  2. B \(\frac{2}{3}\,\left( {x + 2} \right)\)
  3. C \(\frac{2}{3}\,\left( {x - 4} \right)\)
  4. D \(\frac{1}{3}\,\left( {x + 4} \right)\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(\frac{1}{3}\,\left( {x + 4} \right)\)

Step-by-step Solution

Detailed explanation

\(\text { Put } 2 x-1=t^{2}\) \(\Rightarrow \int \frac{x+1}{\sqrt{2 x-1}} d x\) \(=\int\left(\frac{t^{2}+3}{2}\right) d t=\frac{t^{3}}{6}+\frac{3 t}{2}+C\) \(=t\left(\frac{t^{2}}{6}+\frac{3}{2}\right)+C\) \(=\sqrt{2 x-1}\left(\frac{x+4}{3}\right)+C\)
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