Two Fantastic integrals

Exercise:

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

Solution: Let \(u={{x}^{x}}\Leftrightarrow \ln u=x\ln x\Leftrightarrow \frac{du}{u}=\left( \ln x+1 \right)dx\)

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



Exercise:

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

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

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

Now Let $x={{\sin }^{2}}\theta \Rightarrow dx=2\sin \theta \cos \theta \,d\theta $

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

                      $=2\int{\csc \theta \,d\theta -2}\int{d\theta }=2\int{\csc \theta \,d\theta }-2\theta $

But $\int{\csc \theta \,d\theta =}\int{\frac{\csc \theta \left( \csc \theta -\cot \theta  \right)}{\csc \theta -\cot \theta }}\,d\theta =\int{\frac{d\left( \csc \theta -\cot \theta  \right)}{\csc \theta -\cot \theta }}=\ln \left| \csc \theta -\cot \theta  \right|+c$

Hence $\int{\frac{1-\sqrt{x}}{x\sqrt{1-x}}dx=2\ln \left| \frac{1-\sqrt{1-x}}{\sqrt{x}} \right|-2\arcsin \sqrt{x}+c}$