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Integral e^2x/ 9+e^4x, finding the indefinite integral

User Hdiz
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1 Answer

16 votes
16 votes

The given integral expression is :


\int (e^(2x))/(9+e^(4x))dx

Apply u-substitution method :


\begin{gathered} \text{Let u = e}^(2x) \\ \text{Differentiate with respect x} \\ (du)/(dx)=2e^(2x) \\ du=2e^(2x)dx \\ (du)/(2)=e^(2x)dx \\ \text{ So, substitute }e^(2x)dx\text{ = }(du)/(2)\text{ and u = e}^(2x)\text{ in the given integral } \end{gathered}

Thus the integral become :


\begin{gathered} \int (e^(2x))/(9+e^(4x))dx=\int (e^(2x))/(9+e^(2x)e^(2x))dx \\ \int (e^(2x))/(9+e^(4x))dx=\int \frac{1^{}}{9+u\cdot u^{}}(du)/(2) \\ \int (e^(2x))/(9+e^(4x))dx=(1)/(2)\int \frac{1^{}}{9+u^2^{}}du \end{gathered}

Apply the integral substitution : u = 3v


\begin{gathered} u\text{ =3v} \\ \text{differentiate : du =3dv} \\ \text{substitute the value :} \end{gathered}


\begin{gathered} \int (e^(2x))/(9+e^(4x))dx=(1)/(2)\int \frac{1^{}}{9+u^2^{}}du \\ \int (e^(2x))/(9+e^(4x))dx=(1)/(2)\int (3dv)/(9+(3v)^2) \\ \int (e^(2x))/(9+e^(4x))dx=(1)/(2)\int (3dv)/(9+9v^2) \\ \int (e^(2x))/(9+e^(4x))dx=(1)/(2)\int (3dv)/(9(1+v^2)) \\ \int (e^(2x))/(9+e^(4x))dx=(1)/(2)\int (dv)/(3(1+v^2)) \\ \int (e^(2x))/(9+e^(4x))dx=(1)/(6)\int (dv)/((1+v^2)) \end{gathered}

Apply the integral expression :


\int (dx)/(1+x^2)=\arctan x

So, the expression will be :


\begin{gathered} \int (e^(2x))/(9+e^(4x))dx=(1)/(6)\int (dv)/((1+v^2)) \\ \int (e^(2x))/(9+e^(4x))dx=(1)/(6)\arctan (v)+C \end{gathered}

substitute the value of v = u/3


\begin{gathered} \int (e^(2x))/(9+e^(4x))dx=(1)/(6)\arctan (v)+C \\ \int (e^(2x))/(9+e^(4x))dx=(1)/(6)\arctan ((u)/(3))+C \end{gathered}

Now, substitute u = e^2x


\begin{gathered} \int (e^(2x))/(9+e^(4x))dx=(1)/(6)\arctan ((u)/(3))+C \\ \int (e^(2x))/(9+e^(4x))dx=(1)/(6)\arctan ((e^(2x))/(3))+C \end{gathered}

Answer :


\int (e^(2x))/(9+e^(4x))dx=(1)/(6)\arctan ((e^(2x))/(3))+C

User Arivarasan L
by
2.9k points
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