<s>[INST] Evaluating $\int_{0}^{1} \sqrt{1+x^2} \text{ d}x$ I'm learning integral. Here is my homework: 
$$\int_0^1 \sqrt{1+x^2}\;dx$$
I think this problem solve by change $x$ to other variable. Can you tell me how please. (just direction how to solve)
thanks :)
 [/INST] 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:
$$\begin{eqnarray*}
\left. \frac{1}{2}\ln t-\frac{1}{8t^{2}}+\frac{1}{8}t^{2}\right\vert _{1}^{\sqrt{2}+1} &=&\frac{\ln \left( \sqrt{2}+1\right) }{2}-\frac{1}{8\left( \sqrt{2}
+1\right) ^{2}}+\frac{1}{8}\left( \sqrt{2}+1\right) ^{2}-0 \\
&=&\frac{\ln \left( \sqrt{2}+1\right) }{2}+\frac{\sqrt{2}}{2}.
\end{eqnarray*}$$
</s>