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AleksandrR [38]
3 years ago
14

Integrate

rt{1+e^{2x} } dx" alt="e^{4x}\sqrt{1+e^{2x} } dx" align="absmiddle" class="latex-formula"> please show work so i can learn
Mathematics
1 answer:
AnnyKZ [126]3 years ago
3 0

Answer:

(\frac{(1+e^{2x}) ^{\frac{5}{2} } }{{5}} + \frac{(1+e^{2x} )^{\frac{3}{2} } }{{3}} )+C

Step-by-step explanation:

<u><em> Step(i):-</em></u>

Given that the function

                    f(x) = e^{4x} \sqrt{1+e^{2x} }

Now integrating on both sides, we get

                 \int\limits{f(x)} \, dx = \int\limits{e^{4x} \sqrt{1+e^{2x} } dx

                               =    \int\limits{e^{2x} e^{2x} \sqrt{1+e^{2x} } dx

                         

<u><em>Step(ii):-</em></u>

  Let  1 + e^{2x}  = t

           e^{2x}  = t -1  

          2e^{2x}dx = d t

          e^{2x}dx = \frac{1}{2} d t

                = \int\limits{( \sqrt{1+e^{2x} }) e^{2x} e^{2x} dx

                  = \int\limits {\sqrt{t}(t-1)\frac{1}{2} dt }

                 = \frac{1}{2} \int\limits {\sqrt{t} (t) -\sqrt{t} ) dt }

                = \frac{1}{2} \int\limits {(t^{\frac{1}{2}  } t^{1} +t^{\frac{1}{2} } ) } \, dx

                = \frac{1}{2} \int\limits {(t^{\frac{3}{2}  } +t^{\frac{1}{2} } ) } \, dx

               = \frac{1}{2} (\frac{t^{\frac{3}{2} +1} }{\frac{3}{2}+1 } + \frac{t^{\frac{1}{2} +1} }{\frac{1}{2}+1 } )+C

              =  \frac{1}{2} (\frac{t^{\frac{3}{2} +1} }{\frac{5}{2} } + \frac{t^{\frac{1}{2} +1} }{\frac{3}{2} } )+C

             = \frac{1}{2} (\frac{t^{\frac{5}{2} } }{\frac{5}{2} } + \frac{t^{\frac{3}{2} } }{\frac{3}{2} } )+C

            = (\frac{(1+e^{2x}) ^{\frac{5}{2} } }{{5}} + \frac{(1+e^{2x} )^{\frac{3}{2} } }{{3}} )+C

<u><em>Final answer:-</em></u>

= (\frac{(1+e^{2x}) ^{\frac{5}{2} } }{{5}} + \frac{(1+e^{2x} )^{\frac{3}{2} } }{{3}} )+C

             

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