JEE Main 2019 — Indefinite Integration Question with Solution
JEE Main 2019 (11 Jan Shift 2)
Question
If where is a constant
of integration, then is equal to:
Choose an option
Show full solutionCorrect option: D
Correct answer
D
Step-by-step explanation
Let
Put
$\begin{array}{l}
\therefore \quad 2 x-1=t^{2} \Rightarrow d x=t d t \\
I=\int \frac{\left(t^{2}+3\right)}{2} d t=\frac{t^{3}}{6}+\frac{3 t}{2}+C \\
=\frac{(2 x-1)^{\frac{3}{2}}}{6}+\frac{3}{2}(2 x-1)^{\frac{1}{2}}+C \\
=\sqrt{2 x-1}\left(\frac{x+4}{3}\right)+C \\
=f(x) \cdot \sqrt{2 x-1}+C
\end{array}$
Hence,
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This is a previous-year question from JEE Main 2019, covering the Indefinite Integration chapter of Mathematics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.