# Calculate the area of the curve $\cfrac{e^x}{e^{2x}+9}$ between the x-axis

6.Calculate the area located on x-axis and below the curve $$y=\cfrac{e^x}{e^{2x}+9}$$

I've thinking of finding the intersection points of the curve and $$y=0$$

\begin{align} e^x& = 0 \qquad /\ln \\ \ln e^x& = \ln 0 \\ x& = 1\\ \end{align} And then I have :

$$\int_0^1\cfrac{e^x}{e^{2x}+9}$$ I don't know how to find the other point of intersection, it may be finding critical point on $$e^{2x}+9$$, and then using improper integrals?

$$\mathrm{e}^x$$ is always positive. (Looks at its graph, for instance.) Consequently, your quotient for defining $$y$$ is always a ratio of two positive numbers, so $$y$$ never intersects the $$x$$-axis.
In your argument, $$\ln 0$$ is undefined. (The natural logarithm diverges to $$-\infty$$ as its argument decreases to $$0$$.)
This function never crosses the $$x$$ axis, so you're looking for $$\int_{-\infty}^\infty\frac{e^x}{e^{2x}+9}\mathrm dx$$which can be solved by integration by substitution.
Hint: $$e^x>0$$ for all real $$x$$