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Решите задачу:

$\displaystyle\\\int\limits^2_0 {\frac{8x}{\sqrt{1+2x^2}} } \, dx=\{t=\sqrt{1+2x^2}\ \ \ t'=4x*\frac{1}{2\sqrt{1+2x^2}}\}=\\\\\\=8\int\limits^2_0 {\frac{x}{\sqrt{1+2x^2}}*\frac{1}{4x*\dfrac{1}{2\sqrt{1+2x^2}} } } \, dt=8\int\limits^2_0 {\frac{1}{\sqrt{1+2x^2}}*\frac{1}{\dfrac{2}{\sqrt{1+2x^2}} } } \, dt=\\\\\\=8\int\limits^2_0 {\frac{1}{\sqrt{1+2x^2}}*\frac{\sqrt{1+2x^2}}{2} } \, dt=8\int\limits^2_0 {\frac{1}{2} } \, dt=(8*\frac{1}{2}t)=4t=4\sqrt{1+2x^2}\mid^2_0=\\\\\\$

$\displaystyle\\=4\sqrt{1+2*2^2}-4\sqrt{1+2*0}=4\sqrt{1+8}-4\sqrt{1}=4*3-4=12-4=8$