At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this and **thousands** of other questions.

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

Join Brainly to access

this expert answer

SEE EXPERT ANSWER

To see the **expert** answer you'll need to create a **free** account at **Brainly**

or \(x\to-\infty^-\) wither, right?

either*

you gotta start really really far out i guess

http://www.themathpage.com/acalc/infinity.htm

@mathslover no, I think you are thinking of \(x\to+\infty\)

oh ... yes .

I'm still trying to figure out this:
\[\int\limits \frac{x \dot\ \sin x}{1 + \cos^2x} dx\]

\(∞^\pm\) aren't even in the Extended real number line

good point^

me too :)
@zarkon actually taught me how to solve this, but he made use of the bounds heavily

There are many indefinite integrals with solutions that are not finite.

true... I am stretching my capacities here admittedly

Well, if by finite you mean a specific value.