integrate sqrt(x+1) over sqrt(x+2)

In this video, I showed how to use varieous integration technique to integrate a rational radical integrand

[@killer_queen2314]

It's been a while since I've watched one of your videos also sorry I failed your class😅

[@lawrencejelsma8118]

If among all other memory requirements you have to know the integrals of sec(x) and sec^3(x) in a test ... Where does memorizing trig integrals end? 😬 ... then I guess I fail his class too. 😢 Lucky for me I'm not in integral Calculus and have a long list of Integral cheat sheet trig formulas of CRC tables to avoid that problem on my job!! 😂

[@Sg190th]

What if you used hyperbolic sub?

[@naturallyinterested7569]

Then you'd find the -ln|...| in the solution be replaced with -1/2 * arccosh(2x+3)

[@hazwi]

at the end, you can get rid of the absolute value on the natural log since the sum of square roots is always positive

[@DEYGAMEDU]

Sir I have sent you a derivative question in your mail, please solve that

[@The_Bad_Ant]

Awesome explain, thank you teacher 💐💙 31/8/2024 SATURDAY 10:52 AM

[@dougaugustine4075]

Double substitution and a pair of trig identities. Wow.

[@Mephisto707]

The absolute value inside ln in the final answer is unnecessary.

[@FOUADNAJATMA]

❤

[@hosseinmortazavi7903]

Nice professor

[@canyoupoop]

2:55 Do you "SEE" a problem in both of those indefinite integrals?

[@icannotchoose]

This is very well explained! I'd love to see more integral problems like this.

[@atrixiousscramasax6686]

whats sec

[@leonmancaj3690]

Bless you good man

[@Mehrdad_Basiry-fj4rl]

Fantastic question...🎉🎉🎉. Learn new methods...

[@SodiqjonSobirov-w5h]

Thank you teacher

[@zeravam]

I used the sustitution sen^2(theta)=1/(x+2), obtaining: cosec(theta)=sqr(x+2) and cotan(theta)=sqr(x+1). My integral was: (2cosec(theta)-2cosec^3(theta)) I had a equivalent result: ln|sqr(x+2)-sqr(x+1)|+sqr((x+2)(x+1))

[@fantasypvp]

The first thing i thought wheb i saw that integral was to immediately put x+1 = tan^2(u), then the top simplifies, the bottom turns to sec and simplifies, so youve got the integral of sin(u) * dx/du du

[@flamewings3224]

To be honest, I don’t know why am I continuing to learning integration math more… I think I’ve found it funny and interesting since my hight school. So, for last question in the video, we actually can get rid of absolute value inside the natural logarithm, cause inside we have a sum of two square roots, which is always greater than zero, so we don’t need absolute value. But… Is it a mistake to just leave it here?

[@BhaiyaMathsWaale]

Use x+1=t^2 Very easy question

[@jaychirmade]

Can we use the standard formula for √(x^2-a^2)?

[@marcgriselhubert3915]

When you arrive at sqrt(u^2 - 1) better use u = cosh(t), with t >0

[@trinugroho3832]

sir, can you explain how can u^2-1 be sec theta?

[@surendrakverma555]

Thanks Sir for excellent explanation and good handwriting. 😊

[@enejedhddhd6882]

I already memorized radical x^2 - a^2 formula😂 i had an ode exam yesterday

[@joeykraftx9444]

Nice. if we let ch t=u as 2nd substitution i find the resolution much easier.

[@holyshit922]

(x+1)/(x+2) = u^2 This substitution lead us to integrating rational function without any secants After substitution proposed by me integration by parts will simplify partial fraction decomposition

[@PrimeNewtons]

Yes, that works too. Answer looking slightly different.

[@JourneyThroughMath]

As far as the plus/minus is concerned, couldnt you ha e factored it out as a constant and then your final answer is +_answer?

[@JourneyThroughMath]

If Im not much mistaken you can drop the absolute value. Both square roots are greater than 0.

[@Tisakoreann]

AMAZING

[@m.h.6470]

Wolfram Alpha gives a very different result. Looks like your result only works for x > -1, so only for the real numbers. If you want complex solution, your way doesn't work.

[@niloneto1608]

Because the function itself is only defined for x>-2. And usually complexo numbers doesn't associate with integrals of a single variable function.

[@Tomorrow32]

@@niloneto1608Yes, my friend. When you see a problem, think of never dividing by zero. X can approach -2 but never reaches it. X is real by definition.

[@m.h.6470]

The function is undefined at x = -2 and is 0 at x = -1, but x < -2 and -2 < x < -1 can be calculated. The results for x < -2 are complex numbers, but they are still possible. And according to Wolfram Alpha, the integral CAN be calculated, so that these values are included.