Answers to the question of the integral of $\frac {1} {x}$ are all based on an implicit assumption that the upper and lower limits of the integral are both positive real numbers.

The integral which you describe has no closed form which is to say that it cannot be expressed in elementary functions. For example, you can express $\int x^2 \mathrm {d}x$ in elementary …

Feb 17, 2025 · The noun phrase "improper integral" written as $$ \int_a^\infty f (x) \, dx $$ is well defined. If the appropriate limit exists, we attach the property "convergent" to that expression …

Feb 4, 2018 · The integral of 0 is C, because the derivative of C is zero. Also, it makes sense logically if you recall the fact that the derivative of the function is the function's slope, because …

Nov 29, 2013 · Using "indefinite integral" to mean "antiderivative" (which is unfortunately common) obscures the fact that integration and anti-differentiation really are different things in general.

Sep 27, 2013 · My HW asks me to integrate $\sin (x)$, $\cos (x)$, $\tan (x)$, but when I get to $\sec (x)$, I'm stuck.

Dec 15, 2017 · A different type of integral, if you want to call it an integral, is a "path integral". These are actually defined by a "normal" integral (such as a Riemann integral), but path …

Aug 9, 2017 · I've been learning the fundamental theorem of calculus. So, I can intuitively grasp that the derivative of the integral of a given function brings you back to that function. Is this …

If by integral you mean the cumulative distribution function $\Phi (x)$ mentioned in the comments by the OP, then your assertion is incorrect.

Mar 17, 2015 · A different approach, building up from first principles, without using cos or sin to get the identity, $$\arcsin (x) = \int\frac1 {\sqrt {1-x^2}}dx$$ where the integrals is from 0 to z. …