Integral of the Day 10.25.23 | SO SPICY!!! I loved it!!! | Math with Professor V

Get ready: this integral is a SPICY one! I have to be honest: I didn't know how to solve it immediately, and had to play around with it for a few minutes. I love when that happens, and I hope you all enjoy this video and integral as much as I did. Now off to hot yoga sculpt, then work, then ballet! Don't forget to give the video a "LIKE" and comment down below. :)

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Professor V

Calculus 2 Lecture Videos on Integration:

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[georgesadler]

Professor V, thank you for another challenging and interesting Integral of the Day. Professor V, when I first saw this Integral the first thing that I did was to let u equal to e raised to the x which implies that du is equal to e raised to the x dx. After substituting u and du into the original integral, the new integral becomes the square root of (u -1) du/ (u +8). The second substitution is to let t equal to square root of (u -1) which implies that t square is equal to u -1 and 2t dt is equal to du. Professor V, the new integral becomes 2 t squared dt/ t squared plus 9, which is similar yours in the video. After performing the long division and then the Integration, my final solution is 2 times the square root of (e raised to the x -1) - 6 Arc tan ((square root of (e raised to the x -1)/3) + Constant. When I plug in the limits of integration, I end up with 6 - 3Pi/2, which matches your solution. My favorite part of this Integral is the substitutions. This is an error free video/lecture on YouTube TV with Professor V from Math TV.

[DaHoKilla]

I havent seen any of your videos recently but finally got back to it, do you have Calculus 3 or Statics Videos?

[ayoubejraidi]

hi thanks for your videos, here is my answer before watching (6 - 3pi/2). if I were one of your students, could you rate me from 1 to 10.