It can't be zero!

The integral $\displaystyle I = \int_{-1}^1 \sqrt{1+x^2} \ dx$ is an even function. It should be $\displaystyle I = {\color{#D61F06}2} \int_{\color{#D61F06}0}^1 \sqrt{1+x^2} \ dx$ .

$\begin{aligned} I & = \int_{-1}^1 \sqrt{1+x^2} \ dx & \small \color{#3D99F6} \text{As the integrand is even} \\ & = {\color{#3D99F6}2} \int_{\color{#3D99F6}0}^1 \sqrt{1+x^2} \ dx & \small \color{#3D99F6} \text{Let } x = \tan \theta, \ dx = \sec^2 \theta \ d \theta \\ & = 2\int_0^{\pi/4} \sqrt{\color{#3D99F6}1+\tan^2\theta} \sec^2 \theta \ d \theta & \small \color{#3D99F6} \text{Note that }1+\tan^2 \theta = \sec^2 \theta \\ & = 2\int_0^{\pi/4} \sec^3 \theta \ d \theta & \small \color{#3D99F6} \text{Using reduction formula} \\ & = 2 \left(\frac {\sec^{3-2} \theta \tan \theta}{3-1} \bigg|_0^{\pi/4} + \frac {3-2}{3-1}\int_0^{\pi/4} \sec^{3-2} \theta \ d \theta\right) \\ & = \sqrt 2 + \color{#3D99F6}\int_0^{\pi/4} \sec \theta \ d \theta & \small \color{#3D99F6} \text{See note 1 below.} \\ & = \sqrt 2 + \color{#3D99F6} \ln(\tan \theta + \sec \theta)) \bigg|_0^{\pi/4} \\ & = \sqrt 2 + \color{#3D99F6} \ln(1 + \sqrt 2) & \small \color{#3D99F6} \text{See note 2 below.} \\ & = \sqrt 2 + \color{#3D99F6} \sinh^{-1} 1 \end{aligned}$

$\implies m+n = 2+1 = \boxed{3}$

$$

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Note 1:

$\small \begin{aligned} I & = \int \sec x \ dx \\ & = \int \frac {\sec x \tan x + \sec^2 x}{\tan x + \sec x} dx & \color{#3D99F6} \text{Let }u = \tan x + \sec x, \ du = (\sec^2 x + \tan x \sec x) \ dx \\ & = \int \frac 1u du \\ & = \ln u + C & \color{#3D99F6} \text{where }C \text{ is the constant of integration.} \\ & = \ln(\tan x + \sec x) + C \end{aligned}$

Note 2:

$\small \begin{aligned} \sinh \left(\ln(1+\sqrt 2)\right) & = \frac {e^{\ln(1+\sqrt 2)} - e^{-\ln(1+\sqrt 2)}}2 \\ & = \frac {1 + \sqrt 2 - \frac 1{1+\sqrt 2}}2 \\ & = \frac {1 + \sqrt 2 - \frac {\sqrt 2-1}{(\sqrt 2+1)(\sqrt 2-1)}}2 \\ & = \frac {1+\sqrt 2 - \sqrt 2 + 1}2 = 1 \\ \implies \sinh^{-1} 1 & = \ln (1+\sqrt 2) \end{aligned}$

$\displaystyle \large \int_{- 1}^{1} \sqrt{1 + x^2} \ dx$

$\text{Since the function } \sqrt{1+x^2} \ \text{ is even, the above integral can be written as -} \\ I = \displaystyle 2\int_{0}^{1} \sqrt{1 + x^2} \ dx\ \quad\quad\small{{\color{#3D99F6}let \sinh(u) = x \implies \cosh(u)du = dx.}} \\ \implies {I = } \displaystyle 2\int_{x=0}^1 {\color{#3D99F6}\sqrt{1+ \sinh^2(u)}}{\color{#3D99F6}\cosh(u) du} = \displaystyle 2\int_{x=0}^1 \sqrt{\cosh^2(u)}\cosh(u) du. \\ \implies \displaystyle 2\int_{x=0}^1 \cosh^2(u)du \\ \implies \displaystyle I = 2\int_{x=0}^1 \frac{\cosh(2u)+1}{2} du \quad\quad { \small \color{#3D99F6} \bigg|\cosh(2u) = 2\cosh^2(u) - 1 \implies \cosh^2(u) = \small \frac{\cosh(2u)+1}{\color{#3D99F6}2}} \\ \implies I= \displaystyle 2\cdot\frac12\int_{x=0}^1 \cosh(2u) + 1 du = \left(\frac{\sinh(2u)}{2} + u \right){\bigg|_{x=0}^1} \\ \implies I = \displaystyle \left( \frac{2\sinh(u)\cosh(u)}{2} + u \right){\bigg|_{x=0}^1} \quad\quad \small{\color{#3D99F6} \bigg|\sinh(2u) = 2\sinh(u)\cosh(u)} \\ \text{Writing I in terms of x we get -} \\ I = \displaystyle \left({\color{#3D99F6}x} {\color{#3D99F6}\sqrt{1+x^2}} + \sinh^{-1}({\color{#3D99F6}x}) \right){\bigg|_{x=0}^1} \quad\quad\small{\color{#3D99F6} \bigg|\sinh(u) = x, \ \cosh(u) = }{\color{#3D99F6}\sqrt{1+\sinh^2(u)}} \implies {\color{#3D99F6}\cosh(u) = } {\color{#3D99F6}\sqrt{1+x^2}} \\ \implies I = \displaystyle (1\cdot \sqrt{1+1^2} + \sinh^{-1}(1) - 0\cdot \sqrt{1+0^2} - \sinh^{-1}(0)) = \boxed{\displaystyle (\sqrt{2} + \sinh^{-1}(1))} \\ \implies \boxed{\displaystyle I = \sqrt{2} + \sinh^{-1}(1)} \quad\quad \small \color{#3D99F6} \implies m = 2, n = 1 \\ \implies m + n = \boxed3$

$\int { \sqrt { 1+{ x }^{ 2 } } } dx$

Substitute $t=x+\sqrt { 1+{ x }^{ 2 } } \\ x=\frac { { t }^{ 2 }-1 }{ 2t } \\ dx=\frac { { t }^{ 2 }+1 }{ 2{ t }^{ 2 } }$

$\int { \sqrt { 1+{ x }^{ 2 } } } dx$ = $\frac { 1 }{ 4 } \int { \frac { { t }^{ 4 }+{ 2t }^{ 2 }+1 }{ { t }^{ 3 } } } dt\quad =\quad \frac { { t }^{ 2 } }{ 8 } +\frac { ln|t| }{ 2 } -\frac { 1 }{ 8{ t }^{ 2 } } \quad =\quad \frac { { (x+\sqrt { 1+{ x }^{ 2 } } ) }^{ 2 } }{ 8 } +\frac { ln|x+\sqrt { 1+{ x }^{ 2 } } | }{ 2 } -\frac { 1 }{ 8({ x+\sqrt { 1+{ x }^{ 2 } } ) }^{ 2 } }$

So, its value between -1 and 1 is $\sqrt { 2 } +ln(\sqrt { 3+\sqrt { 8 } } )$

Considering $\sqrt { 3+\sqrt { 8 } }$ as $\sqrt { \frac { 3+1 }{ 2 } } +\sqrt { \frac { 3-1 }{ 2 } } =\quad \sqrt { 2 } +1$

The value of the integral is $\sqrt { 2 } +ln(1+\sqrt { 2 } )=\sqrt { 2 } +\sinh ^{ -1 }{ (1) }$

So, m+n=2+1=3