Решить определенный интеграл

$\int \sqrt{1+3x^2}\, dx=\int \frac{1+3x^2}{\sqrt{1+3x^2}} dx=\int \frac{dx}{\sqrt{1+3x^2}} +\int \frac{3x^2}{\sqrt{1+3x^2}} dx=I\\\\a)\; \; \int \frac{dx}{\sqrt{1+3x^2}}dx=[\, t=\sqrt3x\; ,\; dt=\sqrt3\, dx\, ]=\frac{1}{\sqrt3}\int \frac{dt}{1+t^2}=\\\\=\frac{1}{\sqrt3}arctgt+C=\frac{1}{\sqrt3}arctg(\sqrt3x)+C\\\\b)\int \frac{3x\cdot x\, dx}{\sqrt{1+3x^2}}=[\, u=x,\; du=dx\; ,\; dv=\frac{3x\, dx}{\sqrt{1+3x^2}}\; ,\; v=\int \frac{3x\, dx}{\sqrt{1+3x^2}}=\\\\=(t=1+3x^2,\; dt=6x\, dx)=$

$=\frac{1}{2}\int \frac{dt}{\sqrt{t}}=\frac{1}{2}\cdot 2\sqrt{t}+C=\sqrt{1+3x^2}\, ]=x\cdot \sqrt{1+3x^2}-\int \sqrt{1+3x^2}\, dx;\\\\\\I=\int \sqrt{1+3x^2}\, dx=\frac{1}{\sqrt3}arctg(\sqrt3x)+x\cdot \sqrt{1+3x^2}-\int \sqrt{1+3x^2}\, dx\; \Rightarrow \\\\2\int \sqrt{1+3x^2}\, dx=\frac{1}{\sqrt3}arctg{\sqrt3x}+x\cdot \sqrt{1+3x^2}+2C\\\\\int \sqrt{1+3x^2}\, dx=\frac{1}{2\sqrt3}arctg(\sqrt3x)+\frac{x}{2}\cdot \sqrt{1+3x^2}+C$

$\int _0^1\sqrt{1+3x^2}\, dx=\frac{1}{2\sqrt3}arctg(\sqrt3x)|_0^1+$ $\frac{x}{2}\cdot \sqrt{1+3x^2}|_0^1=$

$=\frac{1}{2\sqrt3}arctg\sqrt3+\frac{1}{2}\sqrt{4}=\frac{1}{2\sqrt3}\cdot \frac{\pi}{3}+1=\frac{\pi \sqrt3}{18}+1$