Question

Calculati $\int\limits^1_0 {\sqrt{x^{2}+8 } } \, dx$

Answer 1

$I=\int\limits^1_0 {\sqrt{x^2+8} } \, dx =\int\limits^1_0 {\frac{x^2+8}{\sqrt{x^2+8}} } \, dx =\int\limits^1_0 {\frac{x^2}{\sqrt{x^2+8}} } \, dx+\int\limits^1_0 {\frac{8}{\sqrt{x^2+8}} } \, dx=8\int\limits^1_0 {\frac{1}{\sqrt{x^2+8}} } \, dx+\int\limits^1_0 {x(\sqrt{x^2+8})' } \, dx=8ln|x+\sqrt{x^2+8}|^1_0+x \sqrt{x^2+8}|^1_0-\int\limits^1_0 {x'\sqrt{x^2+8} } \, dx =$

$=8ln|1+\sqrt{1+8}|-8ln |0+\sqrt{0+8}|+ 1*\sqrt{1+8}-0\sqrt{0+8}-I=8(ln4-ln2\sqrt{2} )+3-0-I=>2I=8ln(\frac{4}{2\sqrt{2} } )+3=8ln(\frac{2}{\sqrt{2} } )+3=8ln\sqrt{2} +3=8ln2^{\frac{1}{2} }+3=8(\frac{1}{2}ln2+\frac{3}{8} )=>2I=8(\frac{1}{2}ln2+\frac{3}{8} )=>I=8(\frac{1}{4}ln2+\frac{3}{16} )=>\int\limits^1_0 {\sqrt{x^2+8} } \, dx =8(\frac{1}{4}ln2+\frac{3}{16} )$