Clarification on Simplification of a radical

Question

I recently solved the following integral:

$$ \int _1^{\sqrt{3}}\frac{\sqrt{1+x^2}}{x^2}dx $$

After integrating, I obtained the result:

$$ \frac{1}{2}\ln\frac{2-\sqrt2}{2-\sqrt3}+\frac{1}{2}\ln\frac{2+\sqrt3}{2+\sqrt2}+\sqrt2-\frac{2}{\sqrt3} $$

However, the answer provided in my textbook differs slightly:

$$ \sqrt{2}-\frac{2}{\sqrt{3}}+\log\left(\frac{2+\sqrt{3}}{1+\sqrt{2}}\right) $$

The provided answer in the book clearly looks better and more concise.

I'm particularly interested in understanding how

$$ \sqrt{\frac{(2-\sqrt{2})(2+\sqrt{3})}{(2-\sqrt{3})(2+\sqrt{2})}} $$

is exactly equal to

$$ \frac{2+\sqrt{3}}{1+\sqrt{2}} $$

Could someone kindly provide an explanation for this simplification as it is not obvious to me? Thank you!

Answer 1

Note that$$\frac{2+\sqrt3}{2-\sqrt3}=\frac{\left(2+\sqrt3\right)^2}{\left(2-\sqrt3\right)\left(2+\sqrt3\right)}=\left(2+\sqrt3\right)^2$$and that\begin{align}\frac{2-\sqrt2}{2+\sqrt2}&=\frac{\left(2-\sqrt2\right)\left(2+\sqrt2\right)}{\left(2+\sqrt2\right)^2}\\&=\frac2{\left(2+\sqrt2\right)^2}\\&=\frac{\sqrt2^2}{\left(2+\sqrt2\right)^2}\\&=\frac1{\left(\sqrt2+1\right)^2}.\end{align}

Answer 2

I think we just have to multiply by conjugates, and divide by $\sqrt{2}$ above and below, there we go :

\begin{align} \frac{(2-\sqrt{2})(2+\sqrt{3})}{(2-\sqrt{3})(2+\sqrt{2})} &= \frac{(2-\sqrt{2})(2+\sqrt{3})^{2}}{(2^{2}-\sqrt{3}^{2})(2+\sqrt{2})} \\ &=\frac{(\frac{2}{\sqrt{2}}-1)(2+\sqrt{3})^{2}}{1\times(\frac{2}{\sqrt{2}}+1)} \\ &= \frac{(\sqrt{2}-1)(2+\sqrt{3})^{2}}{\sqrt{2}+1} \\ &= \frac{(\sqrt{2}^{2}-1^{2})(2+\sqrt{3})^{2}}{(1+\sqrt{2})^{2}} \\ &= \frac{(2+\sqrt{3})^{2}}{(1+\sqrt{2})^{2}} \end{align}

And you conclude with the square root.

Answer 3

Textbook Answer

Putting $x\mapsto \frac{1}{x}$ transforms the integral into $$ \begin{aligned} \int_1^{\sqrt{3}} \frac{\sqrt{1+x^2}}{x^2} d x & =-\left[\frac{\sqrt{1+x^2}}{x}\right]_1^{\sqrt{3}}+\int_1^{\sqrt{3}} \frac{1}{x} \frac{x}{\sqrt{11 x^2}} d x \\ & =\sqrt{2}-\frac{2}{\sqrt{3}}+\int_1^{\sqrt{3}} \frac{d x}{\sqrt{1+x^2}} \end{aligned} $$ Putting $x\mapsto \tan \theta$ gives $$ \begin{aligned} \int_1^{\sqrt{3}} \frac{d x}{\sqrt{1+x^2}} = & \int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \frac{1}{\sec \theta} \cdot \sec ^2 \theta d \theta \\ = & {\left[\ln (\sec \theta+\tan \theta \mid]_{\frac{\pi}{4}}^{\frac{\pi}{3}}\right.} \\ = & \ln (2+\sqrt{3})-\ln (\sqrt{2}+1) \\ = & \ln \left|\frac{2+\sqrt{3}}{1+\sqrt{2}}\right| \end{aligned} $$

Answer 4

Noting that $$ \frac{2-\sqrt{2}}{2-\sqrt{3}} \cdot \frac{\sqrt{2}+1}{2+\sqrt{3}}=\sqrt{2} \quad \Rightarrow \quad \frac{2-\sqrt{2}}{3-\sqrt{3}}=\frac{\sqrt{2}(2+\sqrt{3})}{\sqrt{2}+1} $$ Therefore $$ \begin{aligned} \frac{(2-\sqrt{2})(2+\sqrt{3})}{(3-\sqrt{3})(2+\sqrt{2})} & =\frac{\sqrt{2}(2+\sqrt{3})^2}{(\sqrt{2}+1)(2+\sqrt{2})} \\ & =\left(\frac{2+\sqrt{3}}{1+\sqrt{2}}\right)^2 \end{aligned} $$