Verified Solution • 2025–26 Session moderate

QJEEM25Ma785b: JEE Main 2025 Mathematics - Class 12 JEEM25Ma785b Solution: Let . If , then is equal to

Let I(x)=∫dx(x−11)1113(x+15)1513\mathrm{I}(x)=\int \frac{d x}{(x-11)^{\frac{11}{13}}(x+15)^{\frac{15}{13}}} . If I(37)−I(24)=14(1 b113−1c113),b,c∈N\mathrm{I}(37)-\mathrm{I}(24)=\frac{1}{4}\left(\frac{1}{\mathrm{~b}^{\frac{1}{13}}}-\frac{1}{\mathrm{c}^{\frac{1}{13}}}\right), \mathrm{b}, \mathrm{c} \in \mathcal{N} , then 3( b+c)3(\mathrm{~b}+\mathrm{c}) is equal to

Multiple Choice Options

  • A 39
  • B 22
  • C 40
  • D 26

Step-by-Step Verified Solution

amp;I(x)=∫dx(x−11)1113(x+15)1513amp; Put x−11x+15=t⇒26(x+5)2dx=dtamp;I(x)=126∫dtt11/13=126⋅t2/132/13\begin{aligned}& I(x)=\int \frac{d x}{(x-11)^{\frac{11}{13}}(x+15)^{\frac{15}{13}}} \\& \text { Put } \frac{x-11}{x+15}=t \Rightarrow \frac{26}{(x+5)^2} d x=d t \\& I(x)=\frac{1}{26} \int \frac{d t}{t^{11 / 13}}=\frac{1}{26} \cdot \frac{t^{2 / 13}}{2 / 13}\end{aligned}

amp;I(x)=14(x−11x+15)2/13+Camp;I(37)−I(24)=14(2652)2/13−14(1339)2/13amp;=14(122/13−132/13)amp;=14(141/13−191/13)amp;∴b=4,c=9amp;3( b+c)=39\begin{aligned}& \mathrm{I}(\mathrm{x})=\frac{1}{4}\left(\frac{\mathrm{x}-11}{\mathrm{x}+15}\right)^{2 / 13}+\mathrm{C} \\& \mathrm{I}(37)-\mathrm{I}(24)=\frac{1}{4}\left(\frac{26}{52}\right)^{2 / 13}-\frac{1}{4}\left(\frac{13}{39}\right)^{2 / 13} \\& =\frac{1}{4}\left(\frac{1}{2^{2 / 13}}-\frac{1}{3^{2 / 13}}\right) \\& =\frac{1}{4}\left(\frac{1}{4^{1 / 13}}-\frac{1}{9^{1 / 13}}\right) \\& \therefore \mathrm{b}=4, \mathrm{c}=9 \\& 3(\mathrm{~b}+\mathrm{c})=39\end{aligned}

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