Integral Simplification

The integrand is given by: $ \int \frac{(\sqrt{1+x^2}+x)^{10}}{(\sqrt{1+x^2}-x)^9} dx $ We utilize the identity $(\sqrt{1+x^2}+x)(\sqrt{1+x^2}-x) = (1+x^2) - x^2 = 1$. This implies $\sqrt{1+x^2}-x = \frac{1}{\sqrt{1+x^2}+x}$. Substituting this into the integrand simplifies the expression: $ \frac{(\sqrt{1+x^2}+x)^{10}}{\left(\frac{1}{\sqrt{1+x^2}+x}\right)^9} = (\sqrt{1+x^2}+x)^{10} (\sqrt{1+x^2}+x)^9 = (\sqrt{1+x^2}+x)^{19} $ The integral transforms to: $ \int (\sqrt{1+x^2}+x)^{19} dx $

Integration using Substitution

Let $u = \sqrt{1+x^2}+x$. To proceed with the integration, we express $dx$ in terms of $u$ and $du$. We derive $x$ in terms of $u$: From $u = \sqrt{1+x^2}+x$, we get $\sqrt{1+x^2} = u-x$. We also know $\sqrt{1+x^2}-x = \frac{1}{u}$. Subtracting the second relation from the first: $(\sqrt{1+x^2}+x) - (\sqrt{1+x^2}-x) = u - \frac{1}{u}$. This simplifies to $2x = u - \frac{1}{u}$, leading to $x = \frac{1}{2} \left( u - \frac{1}{u} \right)$. Differentiating $x$ with respect to $u$: $ \frac{dx}{du} = \frac{1}{2} \frac{d}{du} \left( u - u^{-1} \right) = \frac{1}{2} (1 - (-1)u^{-2}) = \frac{1}{2} \left( 1 + \frac{1}{u^2} \right) = \frac{u^2+1}{2u^2} $ Thus, $dx = \frac{u^2+1}{2u^2} du$.

Performing the Integral Calculation

Substitute $u$ and the expression for $dx$ into the integral: $ \int u^{19} dx = \int u^{19} \left( \frac{u^2+1}{2u^2} \right) du $ $ = \frac{1}{2} \int \frac{u^{19}(u^2+1)}{u^2} du = \frac{1}{2} \int (u^{19} + u^{17}) du $ Now, perform the integration with respect to $u$: $ = \frac{1}{2} \left( \frac{u^{20}}{20} + \frac{u^{18}}{18} \right) + C $ $ = \frac{u^{20}}{40} + \frac{u^{18}}{36} + C $ Factor out the common term $u^{18}$: $ = u^{18} \left( \frac{u^2}{40} + \frac{1}{36} \right) + C $ To combine the terms inside the parenthesis, find the least common multiple of 40 and 36, which is 360: $ = u^{18} \left( \frac{9u^2}{360} + \frac{10}{360} \right) + C $ $ = \frac{u^{18} (9u^2 + 10)}{360} + C $ $ = \frac{1}{360} u^{18} (9u^2 + 10) + C $

Finding Constants m and n

The integrated result is provided in the form: $ \frac{1}{m} \left( (\sqrt{1+x^2}+x)^n (n\sqrt{1+x^2}-x) \right) + C $ Substitute $u = \sqrt{1+x^2}+x$. The given form transforms to: $ \frac{1}{m} u^n (n\sqrt{1+x^2}-x) + C $ Express the term $(n\sqrt{1+x^2}-x)$ using $u$. We use $x = \frac{1}{2}(u - \frac{1}{u})$ and $\sqrt{1+x^2} = \frac{1}{2}(u + \frac{1}{u})$. $ n\sqrt{1+x^2}-x = n \left( \frac{u+1/u}{2} \right) - \frac{u-1/u}{2} $ $ = \frac{1}{2} \left( n(u+\frac{1}{u}) - (u-\frac{1}{u}) \right) = \frac{1}{2} \left( \frac{nu^2+n - u^2+1}{u} \right) $ $ = \frac{(n-1)u^2 + (n+1)}{2u} $ Substitute this back into the target form: $ \frac{1}{m} u^n \left( \frac{(n-1)u^2 + (n+1)}{2u} \right) + C $ $ = \frac{1}{2m} u^{n-1} ((n-1)u^2 + n+1) + C $ Compare this expression with our calculated result: $ \frac{1}{360} u^{18} (9u^2 + 10) + C $ Matching the powers of $u$, we equate $n-1 = 18$, which gives $n = 19$. Substitute $n=19$ into the expression derived from the target form: $ \frac{1}{2m} u^{18} ((19-1)u^2 + 19+1) + C = \frac{1}{2m} u^{18} (18u^2 + 20) + C $ $ = \frac{1}{2m} u^{18} \cdot 2(9u^2 + 10) + C = \frac{1}{m} u^{18} (9u^2 + 10) + C $ By comparing $\frac{1}{m} u^{18} (9u^2 + 10) + C$ with $\frac{1}{360} u^{18} (9u^2 + 10) + C$, we find the coefficient equality: $ \frac{1}{m} = \frac{1}{360} \implies m = 360 $

Calculating m + n

We have determined that $m=360$ and $n=19$. Both values are natural numbers, as required. The question asks for the sum $m+n$. $ m+n = 360 + 19 = 379 $