4 _0^1(1√3+x^2+√1+x^2) d x-3 _e(√3) is equal to :

Question

$4 \int_0^1\left(\frac{1}{\sqrt{3+x^2}+\sqrt{1+x^2}}\right) d x-3 \log _e(\sqrt{3})$ is equal to :

Accepted Answer

$$\begin{aligned}
& I=4 \int_0^1 \frac{1}{\sqrt{3+x^2}+\sqrt{1+x^2}} d x \
& =2 \int_0^1 \sqrt{3+x^2}-\sqrt{1+x^2} d x
\end{aligned}$$

$$\begin{aligned}
& =2\left[\int_0^1 \sqrt{3+x^2} d x-\int_0^1 \sqrt{1+x^2} d x\right] \
& =2\left[\left(\frac{1}{2} x \sqrt{x^2+3}+\frac{3}{2} \ln \left|\sqrt{3+x^2}+x\right|\right)-\right. \
& \left.\quad\left(\frac{1}{2} x \sqrt{1+x^2}+\frac{1}{2} \ln \left|\sqrt{1+x^2}+x\right|\right)\right]_0^1
\end{aligned}$$

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

Solutions · 1