How do you show whether the improper integral #int e^x/ (e^2x+3)dx# converges or diverges from 0 to infinity?

Answer 1

To determine whether the improper integral

[ \int_{0}^{\infty} \frac{e^x}{e^{2x} + 3} , dx ]

converges or diverges, you can use the limit comparison test.

Let ( f(x) = \frac{e^x}{e^{2x} + 3} ) and ( g(x) = \frac{1}{e^x} ).

First, show that ( f(x) ) and ( g(x) ) are positive, continuous, and decreasing functions for ( x \geq 0 ).

Next, compute the limit:

[ \lim_{x \to \infty} \frac{f(x)}{g(x)} = \lim_{x \to \infty} \frac{\frac{e^x}{e^{2x} + 3}}{\frac{1}{e^x}} ]

[ = \lim_{x \to \infty} \frac{e^{2x}}{e^{2x} + 3} ]

[ = \lim_{x \to \infty} \frac{1}{1 + \frac{3}{e^{2x}}} = 1 ]

Since the limit is a finite positive number, and ( \int_0^\infty g(x) , dx ) converges, by the limit comparison test, ( \int_0^\infty f(x) , dx ) also converges. Therefore, the given integral converges.

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Answer 2

Assuming that the intended integral is #int e^x/(e^(2x)+3) dx# see below.

Integration by substitution will get an arctan whose argument involves #e^x#. As #xrarroo#, the argument will #rarroo#, so arctan #rarr pi/2#
Let #u = e^3# so #du = e^x dx# and the integral becomes
#int 1/(u^2+3) du#
Now let #u = sqrt3 t# to get
#1/sqrt3 int 1/(t^2+1) dt = 1/sqrt3 tan^-1(t)#
Where #t = u/sqrt3 = e^x/sqrt3#
So #int e^x/(e^(2x)+3) dx = 1/sqrt3 tan^-1 (e^x/sqrt3)#

I have omitted the details to properly express the calculation of an improper integral. I assume that the indefinite integral was the difficulty.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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