Let I(x)=∫ d x (x-11)^ (11)/(13) (x+15)^ (15)/(13) . If I(37)- I (24)=(1)/(4)( 1 ~b^ (1)/(13) - 1 c^ (1)/(13) ), b, c ∈ N , then 3(~b+c) is equal to

Correct answer: B. 39 aligned & I(x)=∫ d x (x-11)^ (11)/(13) (x+15)^ (15)/(13) & Put (x-11)/(x+15)=t (26)/((x+5)^2) d x=d t & I(x)=(1)/(26) ∫ d t t^ 1 / 13 =(1)/(26) · t^ 2 / 13 2 / 13 aligned aligned & I(x)=(1)/(4)( x-11 x+15 )^ 2 / 13 +C & I(37)-I(24)=(1)/(4)((26)/(52))^ 2 / 13 -(1)/(4)((13)/(39))^ 2 / 13 & =(1)/(4)( 1 2^ 2 / 13 - 1 3^ 2 / 13 ) & =(1)/(4)( 1 4^ 1 / 13 - 1 9^ 1 / 13 ) & b=4, c=9 & 3( b+c)=39 aligned

JEE Main 202523 JanuaryMorning Shift

Exam: JEE Main
Year: 2025
Date: 23 January
Shift: Morning Shift
Subject: Mathematics
Chapter: Indefinite Integration

JEE Main 2025 · 23 January · Morning Shift · Mathematics · Indefinite Integration

Let I(x)=∫ d x (x-11)^ (11)/(13) (x+15)^ (15)/(13) . If I(37)- I...

Let

\mathrm{I}(x)=\int \frac{d x}{(x-11)^{\frac{11}{13}}(x+15)^{\frac{15}{13}}}

. If

\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 \mathrm{N}

, then

3(\mathrm{~b}+\mathrm{c})

is equal to

Options

A $22$
B $39$
C $40$
D $26$

Correct answer

B. $39$

Step-by-step solution

\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^{1 / 13}}=\frac{1}{26} \cdot \frac{t^{2 / 13}}{2 / 13}\end{aligned}

\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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