integral exercise of $\frac{1+\sqrt{1+{{e}^{x}}}}{{{e}^{x}}+1}$

Exercise:

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

Solution: Let ${{u}^{2}}={{e}^{x}}+1\Leftrightarrow 2udu={{e}^{x}}dx\Leftrightarrow \frac{2u}{{{e}^{x}}}=dx$ but ${{e}^{x}}={{u}^{2}}-1$ hence $dx=\frac{2u}{{{u}^{2}}-1}du$

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

But $\frac{2}{u\left( u-1 \right)}=\frac{A}{u}+\frac{B}{u-1}=\frac{Au-A+Bu}{u\left( u-1 \right)}=\frac{u\left( A+B \right)-A}{u\left( u-1 \right)}$

Hence $A+B=0\,\,\And \,\,-A=2$ that is $B=-A=2$

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

But ${{u}^{2}}={{e}^{x}}+1\Leftrightarrow u=\sqrt{{{e}^{x}}+1}\,\,\,(\because \,{{e}^{x}}+1>0\,\,)$

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