If int frac{(sqrt{1+x^2}+x)^{10}}{(sqrt{1+x^2 | vedclass

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(B) Let $I = \int \frac{(\sqrt{1+x^2}+x)^{10}}{(\sqrt{1+x^2}-x)^9} dx$. Rationalizing the denominator: $I = \int \frac{(\sqrt{1+x^2}+x)^{10}}{(\sqrt{1

Vedclass · Accepted Answer

$379$

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(B) Let $I = \int \frac{(\sqrt{1+x^2}+x)^{10}}{(\sqrt{1+x^2}-x)^9} dx$. Rationalizing the denominator: $I = \int \frac{(\sqrt{1+x^2}+x)^{10}}{(\sqrt{1+x^2}-x)^9} \cdot \frac{(\sqrt{1+x^2}+x)^9}{(\sqrt{1+x^2}+x)^9} dx = \int (\sqrt{1+x^2}+x)^{19} dx$. Let $t = \sqrt{1+x^2}+x$. Then $\frac{dt}{dx} = \frac{x}{\sqrt{1+x^2}} + 1 = \frac{x+\sqrt{1+x^2}}{\sqrt{1+x^2}} = \frac{t}{\sqrt{1+x^2}}$. So,$dx = \frac{\sqrt{1+x^2}}{t} dt$. Since $t = \sqrt{1+x^2}+x$,we have $\sqrt{1+x^2}-x = \frac{1}{t}$. Adding the two equations: $2\sqrt{1+x^2} = t + \frac{1}{t} \implies \sqrt{1+x^2} = \frac{1}{2}(t + \frac{1}{t})$. Substituting this into the integral: $I = \int t^{19} \cdot \frac{1}{2}(t + \frac{1}{t}) \cdot \frac{1}{t} dt = \frac{1}{2} \int (t^{19} + t^{17}) dt$. $I = \frac{1}{2} (\frac{t^{20}}{20} + \frac{t^{18}}{18}) + C = \frac{t^{19}}{2} (\frac{t}{20} + \frac{1}{18t}) + C = \frac{t^{19}}{360} (9t + \frac{10}{t}) + C$. Since $t = \sqrt{1+x^2}+x$ and $\frac{1}{t} = \sqrt{1+x^2}-x$,we have $9t + \frac{10}{t} = 9(\sqrt{1+x^2}+x) + 10(\sqrt{1+x^2}-x) = 19\sqrt{1+x^2} - x$. Thus,$I = \frac{1}{360} ((\sqrt{1+x^2}+x)^{19} (19\sqrt{1+x^2}-x)) + C$. Comparing with the given form,$m = 360$ and $n = 19$. Therefore,$m+n = 360 + 19 = 379$.

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

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