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Sagot :
The given integral expression is :
[tex]\int \frac{e^{2x}}{9+e^{4x}}dx[/tex]Apply u-substitution method :
[tex]\begin{gathered} \text{Let u = e}^{2x} \\ \text{Differentiate with respect x} \\ \frac{du}{dx}=2e^{2x} \\ du=2e^{2x}dx \\ \frac{du}{2}=e^{2x}dx \\ \text{ So, substitute }e^{2x}dx\text{ = }\frac{du}{2}\text{ and u = e}^{2x}\text{ in the given integral } \end{gathered}[/tex]Thus the integral become :
[tex]\begin{gathered} \int \frac{e^{2x}}{9+e^{4x}}dx=\int \frac{e^{2x}}{9+e^{2x}e^{2x}}dx \\ \int \frac{e^{2x}}{9+e^{4x}}dx=\int \frac{1^{}}{9+u\cdot u^{}}\frac{du}{2} \\ \int \frac{e^{2x}}{9+e^{4x}}dx=\frac{1}{2}\int \frac{1^{}}{9+u^2^{}}du \end{gathered}[/tex]Apply the integral substitution : u = 3v
[tex]\begin{gathered} u\text{ =3v} \\ \text{differentiate : du =3dv} \\ \text{substitute the value :} \end{gathered}[/tex][tex]\begin{gathered} \int \frac{e^{2x}}{9+e^{4x}}dx=\frac{1}{2}\int \frac{1^{}}{9+u^2^{}}du \\ \int \frac{e^{2x}}{9+e^{4x}}dx=\frac{1}{2}\int \frac{3dv}{9+(3v)^2} \\ \int \frac{e^{2x}}{9+e^{4x}}dx=\frac{1}{2}\int \frac{3dv}{9+9v^2} \\ \int \frac{e^{2x}}{9+e^{4x}}dx=\frac{1}{2}\int \frac{3dv}{9(1+v^2)} \\ \int \frac{e^{2x}}{9+e^{4x}}dx=\frac{1}{2}\int \frac{dv}{3(1+v^2)} \\ \int \frac{e^{2x}}{9+e^{4x}}dx=\frac{1}{6}\int \frac{dv}{(1+v^2)} \end{gathered}[/tex]Apply the integral expression :
[tex]\int \frac{dx}{1+x^2}=\arctan x[/tex]So, the expression will be :
[tex]\begin{gathered} \int \frac{e^{2x}}{9+e^{4x}}dx=\frac{1}{6}\int \frac{dv}{(1+v^2)} \\ \int \frac{e^{2x}}{9+e^{4x}}dx=\frac{1}{6}\arctan (v)+C \end{gathered}[/tex]substitute the value of v = u/3
[tex]\begin{gathered} \int \frac{e^{2x}}{9+e^{4x}}dx=\frac{1}{6}\arctan (v)+C \\ \int \frac{e^{2x}}{9+e^{4x}}dx=\frac{1}{6}\arctan (\frac{u}{3})+C \end{gathered}[/tex]Now, substitute u = e^2x
[tex]\begin{gathered} \int \frac{e^{2x}}{9+e^{4x}}dx=\frac{1}{6}\arctan (\frac{u}{3})+C \\ \int \frac{e^{2x}}{9+e^{4x}}dx=\frac{1}{6}\arctan (\frac{e^{2x}}{3})+C \end{gathered}[/tex]Answer :
[tex]\int \frac{e^{2x}}{9+e^{4x}}dx=\frac{1}{6}\arctan (\frac{e^{2x}}{3})+C[/tex]
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