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

Question

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?

Answer 1

Hint: $$e^x>0$$ for all real $$x$$

Answer 2

$\mathrm{e}^x$ 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 $0$.)

Answer 3

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.