Integral Exercise asked in 2008 in JEE IIT Exam

Exercise:

Consider $I=\int{\frac{{{e}^{x}}}{{{e}^{4x}}+{{e}^{2x}}+1}dx\,\,\,\,\And \,\,\,J=\int{\frac{{{e}^{-x}}}{{{e}^{-4x}}+{{e}^{-2x}}+1}\,dx}}$

Integrate , $I-J$

Solution: we have $I-J=\int{\left( \frac{{{e}^{x}}}{{{e}^{4x}}+{{e}^{2x}}+1}-\frac{{{e}^{-x}}}{{{e}^{-4x}}+{{e}^{-2x}}+1} \right)\,dx}$

But $\frac{{{e}^{-x}}}{{{e}^{-4x}}+{{e}^{-2x}}+1}=\frac{\frac{1}{{{e}^{x}}}}{\frac{1}{{{e}^{4x}}}+\frac{1}{{{e}^{2x}}}+1}=\frac{\frac{1}{{{e}^{x}}}}{\frac{1+{{e}^{2x}}+{{e}^{4x}}}{{{e}^{4x}}}}=\frac{{{e}^{4x}}}{{{e}^{x}}\left( 1+{{e}^{2x}}+{{e}^{4x}} \right)}=\frac{{{e}^{3x}}}{{{e}^{4x}}+{{e}^{2x}}+1}$

Hence $I-J=\int{\frac{{{e}^{x}}-{{e}^{3x}}}{{{e}^{4x}}+{{e}^{2x}}+1}dx=\int{\frac{{{e}^{x}}\left( 1-{{e}^{2x}} \right)}{{{e}^{4x}}+{{e}^{2x}}+1}dx}}$

Let $u={{e}^{x}}\Leftrightarrow du={{e}^{x}}dx$ So $I-J=\int{\frac{{{e}^{x}}\left( 1-{{e}^{2x}} \right)}{{{e}^{4x}}+{{e}^{2x}}+1}dx}=\int{\frac{1-{{u}^{2}}}{{{u}^{4}}+{{u}^{2}}+1}du}$

But $\frac{1-{{u}^{2}}}{{{u}^{4}}+{{u}^{2}}+1}=\frac{{{u}^{2}}\left( \frac{1}{{{u}^{2}}}-1 \right)}{{{u}^{2}}\left( {{u}^{2}}+1+\frac{1}{{{u}^{2}}} \right)}=\frac{-1+\frac{1}{{{u}^{2}}}}{{{u}^{2}}+\frac{1}{{{u}^{2}}}+1}=-\frac{1-\frac{1}{{{u}^{2}}}}{{{u}^{2}}+\frac{1}{{{u}^{2}}}+1}=-\frac{1-\frac{1}{{{u}^{2}}}}{{{\left( u+\frac{1}{u} \right)}^{2}}-1}$

Remark that ${{u}^{2}}+\frac{1}{{{u}^{2}}}+1={{u}^{2}}+\frac{1}{{{u}^{2}}}+2-1={{\left( u+\frac{1}{u} \right)}^{2}}-1$

So $\int{\frac{1-{{u}^{2}}}{{{u}^{4}}+{{u}^{2}}+1}du}=-\int{\frac{1-\frac{1}{{{u}^{2}}}}{{{\left( u+\frac{1}{u} \right)}^{2}}-1}du}$

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

But $\frac{1}{{{w}^{2}}-1}=\frac{1}{\left( w-1 \right)\left( w+1 \right)}=\frac{A}{w-1}+\frac{B}{w+1}=\frac{Aw+A+Bw-B}{{{w}^{2}}-1}=\frac{\left( A+B \right)w+A-B}{{{w}^{2}}-1}$

So $A+B=0\,\,\And \,\,\,A-B=1\Leftrightarrow A=\frac{1}{2}\,\,\And \,B=-\frac{1}{2}$

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

By backward substitution we get :

$I-J=-\frac{1}{2}\ln \left| \frac{u+\frac{1}{u}-1}{u+\frac{1}{u}+1} \right|+c=-\frac{1}{2}\ln \left| \frac{{{u}^{2}}+1-u}{{{u}^{2}}+1+u} \right|+c=-\frac{1}{2}\ln \left| \frac{{{e}^{2x}}-{{e}^{x}}+1}{{{e}^{2x}}+{{e}^{x}}+1} \right|+c$