Constant for Convergent Improper Integral

In this video, I showed how to manipulate a rational expression of polynomial such that the improper integral will converge.

[@adammohamed5256]

Wonderful!! Needs two cups of tea just to be awake to the end of this wonderful nice explanation.

[@user-wg3lx1mk2s]

Fantastic explanation you’re an amazing teacher thank you!!!

[@Sofia-lh4zi]

great explanation and vibes thank you!!

[@user-iq6qy2bm3e]

This is the best explanation. Thank you sir 🙏🏽

[@christianadlys4473]

Love the engery and passion dude! Keep it up

[@tornikekikacheishvili7318]

Thank you for the good explanation. Can you solve the same problem if we have an integral of x/sqrt (x^2+4) - c/3x+1 ?

[@darcash1738]

What I did was integrate first and use logarithms exponent rule. So I got ln(x^2+1)^1/2 - ln(3x+1)^(1/3*C) I realized that the only way it would even out going to inf is if they were to the same highest power. So, (x^2)^1/2 = x. 1/3*C=1. C = 3. Would that be enough work to show that’s true, or would I need to invoke some law of math at the part where I realized that they need to be of equal highest order 😅

[@skwbusaidi]

After c=3 we should compare to 1/(x^2+1) or 1/(x+1)^2 since the integeral start from 0 And integeral of 1/x^2 from 0 diverges

[@anestismoutafidis4575]

If C is 0, then the integral converges. The exemplary calculation would be Intg Inf->0 x/(x^2+1) •dx Intg x • (x^2+1)=μ^-1•dx Intg x^2/2 • (2x)^-1 • dx Intg inf->0 x^2/4x •dx Inf [x/4 •dx] inf -> 0 by C=0

[@darcash1738]

If c is 0, it becomes 1/2ln(x^2+1)] 0 to inf (can do u sub in your head for these types of problems). Which becomes inf-0 = inf, so it doesn’t converge