Suitengu said:3*integral [ (e^6x)/(1+e^12x)] dx
Im stuck there. I thought I would get to something which I could integrate but I am at a loss.
Suitengu said:I don't think that anything works as I did see that route as well. As a matter of fact the only thing i saw was u = 1 +e^12x but that does work as the derivative of that is 12e^12x which is not in the numerator and from what I see, I doubt I can arrange the numerator to look like that.
Suitengu said:O.O. whoops you are right. then that would be
u = e^6x
du = 6e^6x dx
3*(1/6)*integral [1/(1+u^2)] which is a inverse trig integral.
(1/2)*arctan(u)
(1/2)*arctan(e^6x)
aright thanks man. How do I signal that I have gotten help and found a solution again?
The purpose of integrating 9/(3e^-6x + 3e^6x) is to find the area under the curve represented by the function. This is useful in many fields of science, such as physics, chemistry, and engineering, where the integration of a function can provide valuable information about the behavior of a system.
To solve the integration of 9/(3e^-6x + 3e^6x), you can use the substitution method. Let u = 3e^-6x, then du = -18e^-6x dx. This allows us to rewrite the integral as 9/(-18u + 3u) = 3/(2u), which can be easily integrated. After integrating, substitute back in u to get the final answer.
Yes, there are other methods that can be used to solve this type of integral, such as integration by parts or the use of trigonometric substitutions. However, the substitution method is the most straightforward and efficient method for solving this particular integral.
The limits of integration for 9/(3e^-6x + 3e^6x) will depend on the specific problem you are trying to solve. In general, the limits of integration will be determined by the range of values for x that are relevant to the problem at hand.
The integration of 9/(3e^-6x + 3e^6x) can be applied in many real-world situations, such as calculating the amount of drug in a patient's bloodstream over time, determining the rate of radioactive decay in a sample, or finding the work done by a varying force. In general, any situation where you need to find the area under a curve can be solved using integration.