Let . If , , then 3(b + c) is equal to [2025]

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Solution

Q.
Let $I (x)$ $= \int \frac{d x}{{(x - 11)}^{\frac{11}{13}} {(x + 15)}^{\frac{15}{13}}}$ . If $I (37) - I (24) = \frac{1}{4} (\frac{1}{b^{\frac{1}{13}}} - \frac{1}{c^{\frac{1}{13}}})$ , $b, c \in N$ , then 3(b + c) is equal to [2025]

1 39

2 22

3 40

4 26

Ans.
(1)

Given, $I (x)$ $= \int \frac{d x}{{(x - 11)}^{11 / 13} {(x + 15)}^{15 / 13}}$

Let $\frac{x - 11}{x + 15} = t \Rightarrow \frac{26 d x}{{(x + 15)}^{2}} = d t$

$∴ I (x)$ $= \frac{1}{26} \int \frac{d t}{{(t)}^{11 / 13}}$

$= \frac{1}{26} \times \frac{t^{2 / 13}}{\frac{2}{13}} + C$

$= \frac{1}{4} t^{2 / 13} + C$

$∴ I (x)$ $= \frac{1}{4} {(\frac{x - 11}{x + 15})}^{2 / 13} + C$

$∴ I (37) - I (24) = \frac{1}{4} [{(\frac{1}{2})}^{2 / 13} - {(\frac{1}{3})}^{2 / 13}]$

$\frac{1}{4} (\frac{1}{b^{1 / 13}} - \frac{1}{c^{1 / 13}}) = \frac{1}{4} (\frac{1}{{(4)}^{1 / 13}} - \frac{1}{9^{1 / 13}})$

$∴$ b = 4 and c = 9

$∴$ 3(b + c) = 3(4 + 9) = 39.

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