vault backup: 2025-04-15 09:06:58

education/math/MATH1210 (calc 1)/Integrals.md

@@ -117,3 +117,5 @@ This theorem tells us that a continuous function on the closed interval will obt
 - $\int \dfrac{1}{\sqrt{a^2 - u^2}} du = \arcsin(\frac{u}{a}) +C$ 
 - $\int \dfrac{1}{a^2+u^2}du = \dfrac{1}{a} \arctan(\frac{u}{a}) + C$ 
 - $\int \dfrac{1}{u\sqrt{u^2 - a^2}} du = \dfrac{1}{a}arcsec(\dfrac{|u|}{a}) + C$  
+
+# Length of a Curve