<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] 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 $\int \frac{\mathrm{d} x}{\sqrt{1+x^2}} = \operatorname{arcsinh}(x)$:
$$
  \int_0^1 \sqrt{1+x^2} \mathrm{d} x = \frac{1}{2} \left( \sqrt{2} + \operatorname{arcsinh}(1) \right) = \frac{1}{2} \left( \sqrt{2} + \log(1+\sqrt{2}) \right) \approx 1.1478
$$
</s>