Evaluating $\int_{0}^{1} \sqrt{1+x^2} \text{ d}x$

Integrate by parts to reduce to table integral: $$ \int_0^1 \sqrt{1+x^2} \mathrm{d} x = \left. x \sqrt{1+x^2} \right|_0^1 - \int_0^1 \frac{x^2 {\color\green{+1-1}}}{\sqrt{1+x^2}} \mathrm{d} x = \sqrt{2} - \int_0^1 \sqrt{1+x^2} \mathrm{d} x + \int_0^1 \frac{\mathrm{d} x}{\sqrt{1+x^2}} $$ Now solving the equation for $\int_0^1 \sqrt{1-x^2} \mathrm{d} x$, and using table anti-derivative $\int \frac{\mathrm{d} x}{\sqrt{1+x^2}} = \operatorname{arcsinh}(x)$: $$ \int_0^1 \sqrt{1+x^2} \mathrm{d} x = \frac{1}{2} \left( \sqrt{2} + \operatorname{arcsinh}(1) \right) = \frac{1}{2} \left( \sqrt{2} + \log(1+\sqrt{2}) \right) \approx 1.1478 $$

If we choose to take a (purely) trigonometric route, we might start like this: $$ \eqalign{ x&=\tan\theta\\ dx&=\sec^2\theta\,d\theta\\ I&=\int_0^1\sqrt{1+x^2}\,dx\\ &=\int_0^\frac{\pi}{4}\sec^3\theta\,d\theta\\ } $$ However at this stage, we actually need the substitution $t=\sec\theta+\tan\theta$, believe it or not: $$ \eqalign{ t&=\sec\theta+\tan\theta\\ dt&=\left(\sec\theta\tan\theta+\sec^2\theta\right)\,d\theta =t\sec\theta\,d\theta\\ \frac{dt}{t}&=\sec\theta\,d\theta } $$ And then we need some trigonometry inspirations: $$ \eqalign{ t & = \sec\theta+\tan\theta = \frac{1+\sin\theta}{\cos\theta} = \frac{\cos\theta}{1-\sin\theta} \\ t-\frac1t & = \frac{\cos\theta}{1-\sin\theta} - \frac{\cos\theta}{1+\sin\theta} = 2\tan\theta \\ \tan\theta & = \frac12 \left( t-\frac1t \right) = \frac12 \left( t-t^{-1} \right) \\ \tan^2\theta & = \frac{t^2+t^{-2}}{4} - \frac12 \\ \sec^2\theta & = \frac{t^2+t^{-2}}{4} + \frac12 } $$ We can can then proceed as follows: $$ \eqalign{ I & = \int_0^\frac{\pi}{4}\sec^3\theta\,d\theta = \int_1^{1+\sqrt{2}} \left( \frac{t^2+t^{-2}}{4}+\frac12 \right)\,\frac{dt}{t} \\& = \frac14 \int_1^{1+\sqrt{2}} \left(t+t^{-3}\right)\,dt + \frac12 \int_1^{1+\sqrt{2}} t^{-1}\,dt \\& = \frac18\left[t^2-t^{-2}\right]_1^{1+\sqrt2} + \frac12\left[\ln t \right]_1^{1+\sqrt2} \\& = \frac{\sqrt2+\ln{\left(1+\sqrt2\right)}}{2} } $$ where it is helpful to notice that $$ \left(1+\sqrt2\right)^{-1} = \frac{1}{1+\sqrt2} \cdot \frac{1-\sqrt2}{1-\sqrt2} = \frac{1-\sqrt2}{-1} $$ so that $$ \left(1+\sqrt2\right)^{-2} = \left(1-\sqrt2\right)^2. $$

Since the integrand is a function of $x$ and $\sqrt{ax^{2}+bx+c}$ another option is to use the Euler substitution $\sqrt{ax^{2}+bx+c}=\pm \sqrt{a}x\pm t$, with $a>0$. Choosing $\sqrt{1+x^{2}}=t-x$, squaring both sides and solving for $x$, we obtain $x=\frac{t^{2}-1}{2t}$ and $dx=\frac{ t^{2}+1}{2t^{2}}dt$. The integrand becomes an easily integrable rational fraction of $t$ $$\begin{equation*} \sqrt{1+\left( \frac{t^{2}-1}{2t}\right) ^{2}}\frac{t^{2}+1}{2t^{2}}=\frac{1 }{4}\frac{\left( t^{2}+1\right) ^{2}}{t^{3}}=\frac{1}{2t}+\frac{1}{4t^{3}}+\frac{1}{4}t. \end{equation*}$$ So $$\begin{eqnarray*} \int_{0}^{1}\sqrt{1+x^{2}}dx &=&\int_{1}^{\sqrt{2}+1}\left( \frac{1}{2t}+\frac{1}{4t^{3}}+\frac{t}{4}\right) dt \\ &=&\left. \frac{1}{2}\ln t-\frac{1}{8t^{2}}+\frac{1}{8}t^{2}\right\vert _{1}^{\sqrt{2}+1}. \end{eqnarray*}$$

Added: I've checked the final result:

$$\begin{eqnarray*} \left. \frac{1}{2}\ln t-\frac{1}{8t^{2}}+\frac{1}{8}t^{2}\right\vert _{1}^{\sqrt{2}+1} &=&\frac{\ln \left( \sqrt{2}+1\right) }{2}-\frac{1}{8\left( \sqrt{2} +1\right) ^{2}}+\frac{1}{8}\left( \sqrt{2}+1\right) ^{2}-0 \\ &=&\frac{\ln \left( \sqrt{2}+1\right) }{2}+\frac{\sqrt{2}}{2}. \end{eqnarray*}$$