JEE Main 2025 — Indefinite Integration Question with Solution
JEE Main 2025 (4 Apr Shift 2)
Question
If
is the constant of integration and , then is equal to
is the constant of integration and , then is equal to
Enter your answer
Show full solutionCorrect answer: 379
Correct answer
379
Step-by-step explanation
rationalise
$\begin{aligned}
& \Rightarrow \int \frac{\left(\sqrt{1+x^2}+x\right)^{10}}{\left(\sqrt{1+x^2}-x\right)^9} \times \frac{\left(\sqrt{1+x^2}+x\right)^9}{\left(\sqrt{1+x^2}+x\right)^9} d x \\ & \Rightarrow \int \frac{\left(\sqrt{1+x^2}+x\right)^{19}}{1} d x
\end{aligned}\sqrt{1+\mathrm{x}^2}+\mathrm{x}=\mathrm{t}\left(\frac{\mathrm{x}}{\sqrt{1+\mathrm{x}^2}}+1\right) \mathrm{dx}=\mathrm{dt}\mathrm{dx}=\frac{\mathrm{dt}}{\mathrm{t}} \sqrt{1+\mathrm{x}^2}\sqrt{1+\mathrm{x}^2}+\mathrm{x}=\mathrm{t}\begin{aligned}
& \text { so } \sqrt{1+\mathrm{x}^2}-\mathrm{x}=\frac{1}{\mathrm{t}} \\ & \therefore \sqrt{1+\mathrm{x}^2}=\frac{1}{2}\left(\mathrm{t}+\frac{1}{\mathrm{t}}\right)
\end{aligned}I=\int \mathrm{t}^{19} \cdot \frac{\mathrm{dt}}{\mathrm{t}} \cdot \frac{1}{2}\left(\mathrm{t}+\frac{1}{\mathrm{t}}\right)\begin{aligned}
& \Rightarrow \frac{1}{2} \int\left(\mathrm{t}^{19}+\mathrm{t}^{17}\right) \mathrm{dt} \\ & =\frac{1}{2}\left(\frac{\mathrm{t}^{20}}{20}+\frac{\mathrm{t}^{18}}{18}\right)+\mathrm{C} \\ & =\frac{\mathrm{t}^{19}}{360}\left[9 \mathrm{t}+\frac{10}{\mathrm{t}}\right]+\mathrm{C} \\ & =\frac{\mathrm{t}^{19}}{360}\left[9\left(\mathrm{t}+\frac{1}{\mathrm{t}}\right)+\frac{1}{\mathrm{t}}\right]+\mathrm{C} \\ & \Rightarrow \frac{\left(\sqrt{1+\mathrm{x}^2}+\mathrm{x}\right)^{19}}{360}\left[9\left(2 \sqrt{1+\mathrm{x}^2}\right)+\left(\sqrt{1+\mathrm{x}^2}-\mathrm{x}\right)\right]+\mathrm{C} \\ & \Rightarrow \frac{\left(\sqrt{1+\mathrm{x}^2}+\mathrm{x}\right)^{19}}{360}\left[19 \sqrt{1+\mathrm{x}^2}-\mathrm{x}\right]+\mathrm{C} \\ & \therefore \mathrm{~m}=360, \mathrm{n}=19 \\ & \therefore \mathrm{~m}+\mathrm{n}=379
\end{aligned}$
$\begin{aligned}
& \Rightarrow \int \frac{\left(\sqrt{1+x^2}+x\right)^{10}}{\left(\sqrt{1+x^2}-x\right)^9} \times \frac{\left(\sqrt{1+x^2}+x\right)^9}{\left(\sqrt{1+x^2}+x\right)^9} d x \\ & \Rightarrow \int \frac{\left(\sqrt{1+x^2}+x\right)^{19}}{1} d x
\end{aligned}\sqrt{1+\mathrm{x}^2}+\mathrm{x}=\mathrm{t}\left(\frac{\mathrm{x}}{\sqrt{1+\mathrm{x}^2}}+1\right) \mathrm{dx}=\mathrm{dt}\mathrm{dx}=\frac{\mathrm{dt}}{\mathrm{t}} \sqrt{1+\mathrm{x}^2}\sqrt{1+\mathrm{x}^2}+\mathrm{x}=\mathrm{t}\begin{aligned}
& \text { so } \sqrt{1+\mathrm{x}^2}-\mathrm{x}=\frac{1}{\mathrm{t}} \\ & \therefore \sqrt{1+\mathrm{x}^2}=\frac{1}{2}\left(\mathrm{t}+\frac{1}{\mathrm{t}}\right)
\end{aligned}I=\int \mathrm{t}^{19} \cdot \frac{\mathrm{dt}}{\mathrm{t}} \cdot \frac{1}{2}\left(\mathrm{t}+\frac{1}{\mathrm{t}}\right)\begin{aligned}
& \Rightarrow \frac{1}{2} \int\left(\mathrm{t}^{19}+\mathrm{t}^{17}\right) \mathrm{dt} \\ & =\frac{1}{2}\left(\frac{\mathrm{t}^{20}}{20}+\frac{\mathrm{t}^{18}}{18}\right)+\mathrm{C} \\ & =\frac{\mathrm{t}^{19}}{360}\left[9 \mathrm{t}+\frac{10}{\mathrm{t}}\right]+\mathrm{C} \\ & =\frac{\mathrm{t}^{19}}{360}\left[9\left(\mathrm{t}+\frac{1}{\mathrm{t}}\right)+\frac{1}{\mathrm{t}}\right]+\mathrm{C} \\ & \Rightarrow \frac{\left(\sqrt{1+\mathrm{x}^2}+\mathrm{x}\right)^{19}}{360}\left[9\left(2 \sqrt{1+\mathrm{x}^2}\right)+\left(\sqrt{1+\mathrm{x}^2}-\mathrm{x}\right)\right]+\mathrm{C} \\ & \Rightarrow \frac{\left(\sqrt{1+\mathrm{x}^2}+\mathrm{x}\right)^{19}}{360}\left[19 \sqrt{1+\mathrm{x}^2}-\mathrm{x}\right]+\mathrm{C} \\ & \therefore \mathrm{~m}=360, \mathrm{n}=19 \\ & \therefore \mathrm{~m}+\mathrm{n}=379
\end{aligned}$
Practice this on the real CBT interface
Solve this JEE Main question (and the rest of the Indefinite Integration chapter) on PrepSharp's TCS iON-style CBT player — with timer, bookmarks and session analytics.
Solve interactively →About this question
This is a previous-year question from JEE Main 2025, 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.