Revisiting this integral from the MIT integration Bee

In this video, am revisiting the MIT integration bee problem that Natal ( ‪@youngmathematician9154‬ ) gave me, and am using a different kind of substitution, which i think is easier than what i did in the last video.
My Video with Natal: • An MIT Integration bee...
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[@youngmathematician9154]

That's way easier! I honestly would not have thought of that.

[@ThePhysicsMathsWizard]

😊 Thanks

[@UberHummus]

When I saw a 12 minute video, I knew there had to be a smoother way you had done this. Nice catching the clever substitution here!

[@ThePhysicsMathsWizard]

😊 surely!

[@ramakantpawara]

Really fantastic sar

[@silentintegrals9104]

👏 cool video! integration is art!

[@nicogehren6566]

very nice question

[@ThePhysicsMathsWizard]

Thanks

[@AdAdra621]

When I first saw this question before I clicked to watch your video. I immediately thought substitution was the right way to solve it where u=sqrt(x+sqrt(pow(x,2)+1)) and I solved it and then clicked your video to see your work. It's cool, also substitution but with a much different approach. Although you would have made it much easier if you had converted the equation which relates x to y to separate fractions by dividing each term in the numerator by pow(y,2) before differentiating. You wouldn't even have used the quotient rule you used there

[@ThePhysicsMathsWizard]

Thanks Aaron 😊.
You're genius! The first time I tried to solve it, the substitution that came to my mind was x = tan©, it worked though it was longer.
Nice time buddy!

[@AdAdra621]

@@ThePhysicsMathsWizard Kindly send me a link to that video. I tried trig substitution because you had mentioned it but I am getting stuck on intrg(sqrt(tan©+sec©)sec²©d©)

[@ThePhysicsMathsWizard]

Oh, ok. here: ruclips.net/video/rxMebmgA4hw/видео.html

[@damejelyas]

Noice

[@cookieman2028]

very simple

[@ThePhysicsMathsWizard]

Yeah, thanks for the suggestion 👍

[@babagaming4943]

Okay yaa that is quite a easier way