Answer:
[tex](\frac{(1+e^{2x}) ^{\frac{5}{2} } }{{5}} + \frac{(1+e^{2x} )^{\frac{3}{2} } }{{3}} )+C[/tex]
Step-by-step explanation:
Step(i):-
Given that the function
f(x) = [tex]e^{4x} \sqrt{1+e^{2x} }[/tex]
Now integrating on both sides, we get
[tex]\int\limits{f(x)} \, dx = \int\limits{e^{4x} \sqrt{1+e^{2x} } dx[/tex]
= [tex]\int\limits{e^{2x} e^{2x} \sqrt{1+e^{2x} } dx[/tex]
Step(ii):-
Let [tex]1 + e^{2x} = t[/tex]
[tex]e^{2x} = t -1[/tex]
[tex]2e^{2x}dx = d t[/tex]
[tex]e^{2x}dx = \frac{1}{2} d t[/tex]
= [tex]\int\limits{( \sqrt{1+e^{2x} }) e^{2x} e^{2x} dx[/tex]
= [tex]\int\limits {\sqrt{t}(t-1)\frac{1}{2} dt }[/tex]
= [tex]\frac{1}{2} \int\limits {\sqrt{t} (t) -\sqrt{t} ) dt }[/tex]
= [tex]\frac{1}{2} \int\limits {(t^{\frac{1}{2} } t^{1} +t^{\frac{1}{2} } ) } \, dx[/tex]
[tex]= \frac{1}{2} \int\limits {(t^{\frac{3}{2} } +t^{\frac{1}{2} } ) } \, dx[/tex]
= [tex]\frac{1}{2} (\frac{t^{\frac{3}{2} +1} }{\frac{3}{2}+1 } + \frac{t^{\frac{1}{2} +1} }{\frac{1}{2}+1 } )+C[/tex]
= [tex]\frac{1}{2} (\frac{t^{\frac{3}{2} +1} }{\frac{5}{2} } + \frac{t^{\frac{1}{2} +1} }{\frac{3}{2} } )+C[/tex]
= [tex]\frac{1}{2} (\frac{t^{\frac{5}{2} } }{\frac{5}{2} } + \frac{t^{\frac{3}{2} } }{\frac{3}{2} } )+C[/tex]
= [tex](\frac{(1+e^{2x}) ^{\frac{5}{2} } }{{5}} + \frac{(1+e^{2x} )^{\frac{3}{2} } }{{3}} )+C[/tex]
Final answer:-
= [tex](\frac{(1+e^{2x}) ^{\frac{5}{2} } }{{5}} + \frac{(1+e^{2x} )^{\frac{3}{2} } }{{3}} )+C[/tex]
