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

Question

$4\displaystyle\int_0^1\frac{dx}{\sqrt{3+x^2}+\sqrt{1+x^2}} - 3\log_e(\sqrt3)$ is equal to:

Accepted Answer

Rationalizing, $\dfrac{1}{\sqrt{3+x^2}+\sqrt{1+x^2}} = \dfrac{\sqrt{3+x^2}-\sqrt{1+x^2}}{2}$, so

$$I = 2\int_0^1\left(\sqrt{3+x^2}-\sqrt{1+x^2}\right)dx$$

Using $\int\sqrt{x^2+a^2}\,dx = \dfrac{x}2\sqrt{x^2+a^2}+\dfrac{a^2}2\ln\left(x+\sqrt{x^2+a^2}\right)$:

$$I = 2\left[\left(1+\frac32\ln(1+\sqrt3)-\frac32\ln\sqrt3\right) - \left(\frac{\sqrt2}2+\frac12\ln(\sqrt2+1)\right)\right]$$

Simplifying gives $I = 3\ln\sqrt3+2-\sqrt2-\ln(\sqrt2+1)$, so

$$I - 3\log_e(\sqrt3) = 2-\sqrt2-\log_e(1+\sqrt2)$$

Solutions · 1