If where C is the constant of integration and m, n N, then m + n is equal to __________. [2025]

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Solution

Q.
If $\int \frac{{(\sqrt{1 + x^{2}} + x)}^{10}}{{(\sqrt{1 + x^{2}} - x)}^{9}} d x = \frac{1}{m} ({(\sqrt{1 + x^{2}} + x)}^{n} (n \sqrt{1 + x^{2}} - x)) + C$ where C is the constant of integration and m, n $\in$ N, then m + n is equal to __________. [2025]

Ans.
(379)

Let $I = \int \frac{{(\sqrt{1 + x^{2}} + x)}^{10}}{{(\sqrt{1 + x^{2}} - x)}^{9}} d x = \int \frac{{(\sqrt{1 + x^{2}} + x)}^{19}}{1} d x$ (On rationalising)

Let $\sqrt{1 + x^{2}} + x = t$

$\Rightarrow (\frac{x}{\sqrt{1 + x^{2}}} + 1) d x = d t$

$\Rightarrow d x = \frac{d t}{t} (\sqrt{1 + x^{2}})$

$= \frac{d t}{t} \cdot (\frac{t^{2} + 1}{2 t}) = \frac{t^{2} + 1}{2 t^{2}} \cdot d t$

So, $I = \int t^{19} (\frac{t^{2} + 1}{2 t^{2}}) d t$

$= \frac{1}{2} \int (t^{19} + t^{17}) d t = \frac{1}{2} [\frac{t^{20}}{20} + \frac{t^{18}}{18}] + C$

$= \frac{t^{19}}{360} [9 t + \frac{10}{t}] + C = \frac{t^{19}}{360} [9 (t + \frac{1}{t}) + \frac{1}{t}]$

$= \frac{{(\sqrt{1 + x^{2}} + x)}^{19}}{360} [9 (2 \sqrt{1 + x^{2}}) + (\sqrt{1 + x^{2}} - x)] + C$

$= \frac{{(\sqrt{1 + x^{2}} + x)}^{19}}{360} [19 \sqrt{1 + x^{2}} - x] + C$

Then, m = 360, n = 19

$∴$ m + n = 360 + 19 = 379.

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