Question

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

Answer 1

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`