4 _1^0(1√3+x^2+√1+x^2)dx-3 _e(√3) is equal to :

Question

$4{\int }_{1}^{0}(\frac{1}{\sqrt{3+{x}^{2}}+\sqrt{1+{x}^{2}}})dx-3{\log }_{e}(\sqrt{3})$ is equal to :

Accepted Answer

$\begin{gathered} I=4{\int }_{1}^{0}\frac{1}{\sqrt{3+{x}^{2}}+\sqrt{1+{x}^{2}}}dx \ =2{\int }_{1}^{0}\sqrt{3+{x}^{2}}-\sqrt{1+{x}^{2}}dx \end{gathered}$

$\begin{gathered} =2[{\int }_{1}^{0}\sqrt{3+{x}^{2}}dx-{\int }_{1}^{0}\sqrt{1+{x}^{2}}dx] \ =2[(\frac{1}{2}x\sqrt{{x}^{2}+3}+\frac{3}{2}\ln |\sqrt{3+{x}^{2}}+x|)- \ {\, (\frac{1}{2}x\sqrt{1+{x}^{2}}+\frac{1}{2}\ln |\sqrt{1+{x}^{2}}+x|)]}_{1}^{0} \end{gathered}$

$\begin{gathered} =2[(1+\frac{3}{2}\ln 3-\frac{3}{2}\ln \sqrt{3})-(\frac{\sqrt{2}}{2}+\frac{1}{2}\ln (\sqrt{2}+1))] \ =2(1+\frac{3}{4}\ln 3-\frac{1}{\sqrt{2}}-\frac{1}{2}\ln (\sqrt{2}+1)) \ =3\ln \sqrt{3}+2-\sqrt{2}-\ln (\sqrt{2}+1) \ I-3\ln \sqrt{3}=2-\sqrt{2}-\ln (\sqrt{2}+1) \end{gathered}$

Solutions · 1