integral exercise asked by Vulvaal Pal in Imad Zak math group

Exercise:

Integrate, $\int{\frac{dx}{x+\sqrt{2}\sqrt{{{x}^{2}}-1}}}$

Solution: we have $\frac{1}{x+\sqrt{2}\sqrt{{{x}^{2}}-1}}\times \frac{x-\sqrt{2}\sqrt{{{x}^{2}}-1}}{x-\sqrt{2}\sqrt{{{x}^{2}}-1}}=\frac{x-\sqrt{2}\sqrt{{{x}^{2}}-1}}{{{x}^{2}}-2\left( {{x}^{2}}-1 \right)}=\frac{x-\sqrt{2}\sqrt{{{x}^{2}}-1}}{2-{{x}^{2}}}$

So $\int{\frac{dx}{x+\sqrt{2}\sqrt{{{x}^{2}}-1}}=\int{\frac{x-\sqrt{2}\sqrt{{{x}^{2}}-1}}{2-{{x}^{2}}}dx}}=\int{\frac{x}{2-{{x}^{2}}}dx-\int{\frac{\sqrt{2}\sqrt{{{x}^{2}}-1}}{2-{{x}^{2}}}dx}}$

Let $u=2-{{x}^{2}}\Leftrightarrow du=-2xdx\Leftrightarrow -\frac{du}{2}=xdx$

So $\int{\frac{x}{2-{{x}^{2}}}dx}=-\frac{1}{2}\int{\frac{d\left( 2-{{x}^{2}} \right)}{2-{{x}^{2}}}=-\frac{1}{2}\ln \left| 2-{{x}^{2}} \right|+c}$

We have also $\int{\frac{\sqrt{2}\sqrt{{{x}^{2}}-1}}{2-{{x}^{2}}}dx}=\sqrt{2}\int{\frac{\sqrt{{{x}^{2}}-1}}{2-{{x}^{2}}}dx}=\sqrt{2}\int{\frac{{{x}^{2}}-1}{\left( 2-{{x}^{2}} \right)\sqrt{{{x}^{2}}-1}}dx}$

Let $x=\cosh \theta \Leftrightarrow dx=\sinh \theta d\theta $

So $\int{\frac{{{x}^{2}}-1}{\left( 2-{{x}^{2}} \right)\sqrt{{{x}^{2}}-1}}dx=\int{\frac{{{\cosh }^{2}}\theta -1}{\left( 2-{{\cosh }^{2}}\theta  \right)\sqrt{{{\cosh }^{2}}\theta -1}}\sinh \theta d\theta }}=\int{\frac{{{\sinh }^{2}}\theta }{2-{{\cosh }^{2}}\theta }d\theta }$

$=\int{\frac{{{\sinh }^{2}}\theta }{1+1-{{\cosh }^{2}}\theta }d\theta }=\int{\frac{{{\sinh }^{2}}\theta }{1-{{\sinh }^{2}}\theta }d\theta }=\int{\frac{-1+{{\sinh }^{2}}\theta +1}{1-{{\sinh }^{2}}\theta }d\theta }=\int{\frac{-\left( 1-{{\sinh }^{2}}\theta  \right)+1}{1-{{\sinh }^{2}}\theta }d\theta }$

\[=\int{\left( \frac{1}{1-{{\sinh }^{2}}\theta }-1 \right)d\theta }=\int{\left( \frac{1}{{{\cosh }^{2}}\theta -2{{\sinh }^{2}}\theta }-1 \right)d\theta }=\int{\left( \frac{1}{{{\cosh }^{2}}\theta \left( 1-2{{\tanh }^{2}}\theta  \right)}-1 \right)d\theta }\]

$=\int{\left( \frac{{{\operatorname{sech}}^{2}}\theta }{1-2{{\tanh }^{2}}\theta }-1 \right)d\theta }=\int{\frac{{{\operatorname{sech}}^{2}}\theta }{1-{{\tanh }^{2}}\theta }d\theta -\theta +c}=\int{\frac{d\left( \tanh \theta  \right)}{1-2{{\tanh }^{2}}\theta }-\theta +c}$

Observe that , $\int{\frac{du}{1-2{{u}^{2}}}=\int{\frac{du}{1-{{\left( \sqrt{2}u \right)}^{2}}}}}$ Let $u=\frac{w}{\sqrt{2}}\Leftrightarrow du=\frac{1}{\sqrt{2}}dw$

So $\int{\frac{du}{1-{{\left( \sqrt{2}u \right)}^{2}}}=\frac{1}{\sqrt{2}}\int{\frac{dw}{1-{{w}^{2}}}=\frac{1}{\sqrt{2}}{{\tanh }^{-1}}\left( w \right)+c=\frac{1}{\sqrt{2}}}{{\tanh }^{-1}}\left( \sqrt{2}u \right)+c}$

Hence $\int{\frac{d\left( \tanh \theta  \right)}{1-2{{\tanh }^{2}}\theta }d\theta }=\frac{1}{\sqrt{2}}{{\tanh }^{-1}}\left( \sqrt{2}\tanh \theta  \right)+c$

Observe that $x=\frac{{{e}^{\theta }}+{{e}^{-\theta }}}{2}\Leftrightarrow 2x={{e}^{\theta }}+{{e}^{-\theta }}\Leftrightarrow 2x{{e}^{\theta }}={{e}^{2\theta }}+1\Leftrightarrow {{e}^{2\theta }}-2x{{e}^{\theta }}+1=0$

$\Leftrightarrow {{e}^{2\theta }}-2\left( x \right){{e}^{\theta }}+{{x}^{2}}-{{x}^{2}}+1=0\Leftrightarrow {{\left( {{e}^{\theta }}-x \right)}^{2}}={{x}^{2}}-1\Leftrightarrow {{e}^{\theta }}-x=\sqrt{{{x}^{2}}-1}$

So $\theta =\ln \left( x+\sqrt{{{x}^{2}}-1} \right)$ & $\tanh \theta =\frac{\sinh \theta }{\cosh \theta }=\frac{\sinh \theta }{x}=\frac{\sqrt{{{\cosh }^{2}}\theta -1}}{x}=\frac{\sqrt{{{x}^{2}}-1}}{x}$

So $\int{\frac{dx}{x+\sqrt{2}\sqrt{{{x}^{2}}-1}}=-\frac{1}{2}\ln \left| 2-{{x}^{2}} \right|-{{\tanh }^{-1}}\left( \sqrt{2}\frac{\sqrt{{{x}^{2}}-1}}{x} \right)+\sqrt{2}\ln \left( x+\sqrt{{{x}^{2}}-1} \right)+c}$