integral of sqrt(1+x^2)/x vs integral of x/sqrt(1+x^2)

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NOTE: it's possible to do u sub for the integral of sqrt(1+x^2)/x as well. Thanks to usuario0002hotmail for pointing out.

, trig substitution, calculus 2 tutorial.
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#blackpenredpen #math #calculus #apcalculus

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I used u=sqrt(1+x^2) -> x=sqrt(u^2-1) on the left side integral, and I got the answer after some division :P

hc_

My favourite math guy on YouTube with his spherical microphone.
Love it.

quahntasy

You always save me from my calculus homework, thank you so much!

abigailmedeltoxtle

Thank you so much ❤️❤️❤️
You are my most favourite YouTuber right now!!!

NikitaNair

I really love you YOU SAVED MY LIFE!!!

hacci

I've only had an introductory course on integration. We weren't taught any of the common integration techniques like u-sub, integration by parts, partial fractions, trig sub, you name it. We only had integrals like the integral of x/(1-x^2) which we did by the definition of the integral (you'll even see me employ this kind of thinking in this answer). Anyway, my point is that I haven't yet learned any of this trig sub business so I went the algebraic way. Here's how I did the integral on the left:

∫ √(1 + x²) / x dx

U-sub:
u = √(1 + x²)
u² = 1 + x²
x² = u² - 1
x = √(u² - 1)
dx = 1 / 2√(u² - 1) * 2udu
dx = u / √(u² - 1) du

∫ √(1 + x²) / x dx
= ∫ (u / √(u² - 1)) * (u / √(u² - 1)) du
= ∫ (u / √(u² - 1))² du
= ∫ u² / (u² - 1) du
= ∫ (u² - 1 + 1) / (u² - 1) du
= ∫ (u² - 1) / (u² - 1) + 1 / (u² - 1) du
= ∫ 1 + 1 / (u² - 1) du
= u + ∫ 1 / (u² - 1) du
= u - ∫ 1 / -(u² - 1) du
= u - ∫ 1 / (1 - u²) du
= u - ∫ (1 - u + u) / (1 - u²) du
= u - ∫ (1 - u + u) / ((1 - u)(1 + u)) du
= u - ∫ (1 - u) / ((1 - u)(1 + u)) + u / ((1 - u)(1 + u)) du
= u - ∫ 1 / (1 + u) + u / ((1 - u)(1 + u)) du
= u - ln|1+ u| - ∫ u / ((1 - u)(1 + u)) du
= u - ln|1+ u| - ∫ u / (1 - u²) du

We wanna have the nominator be multiplied by -2 -- you'll soon see why. In order not to change the question, let's also multiply the whole integral by -1/2:
= u - ln|1 + u| - (-1/2) * ∫ -2u / (1 - u²) du

Now you see, the nominator is the derivative of the denominator inside the integral. Recall that d/dx ln(f(x)) = f'(x) / f(x).

= u - ln|1 + u| - (-1/2) * ln|1 - u²| + C
= u - ln|1 + u| + 1/2 * ln|1 - u²| + C

Let's convert the answer back into terms of x:
= √(1 + x²) - ln|1 + √(1 + x²)| + 1/2 * ln|1 - (1 + x²)| + C

The function inside the first natural logarithm is always positive so we may remove the absolute value signs from there:
= √(1 + x²) - ln(1 + √(1 + x²)) + 1/2 * ln|1 - 1 - x²| + C
= √(1 + x²) - ln(1 + √(1 + x²)) + 1/2 * ln|-x²| + C
= √(1 + x²) - ln(1 + √(1 + x²)) + 1/2 * ln(x²) + C

Minding that √(x²) = |x|, let's simplify the second natural logarithm:
= √(1 + x²) - ln(1 + √(1 + x²)) + ln|x| + C
= √(1 + x²) + ln|x| - ln(1 + √(1 + x²)) + C
= √(1 + x²) + ln(|x| / (1 + √(1 + x²))) + C

This is essentially the same function that bprp answers with, you may check yourself.

theimmux

I like U world cause it can also transform into V world and W world and those are my favorites. Theta world will get messy if you take it to another world

mcmage

I like the U-world best. The best part of this video is that the pen fell on the floor twice.

odinfeidje-baug

thank you for this, really well explained :D

Ani

You've always been my life saver in exams

sheikhsehar

Thank you so much . Earlier i was doing it with 1+x²= t method . Thanks for teaching me this ❤️❤️

kanchanmoon

Love your videos, thanks for making them!

peterdhaile

You have the BEST explanations!! Thank you so much for all of the math help ^_^

morganvoissem

Thanks for allowing us to watch an integral battle turn into a three markers battle!

I don't prefer any world over the other. I enter them when I assume they might help me solve an integral.
Btw: The theta world also helps for the integral on the right side.

ralfbodemann