I usually do this method only but just write it out differently because I find that way more intuitive. Instead of directly write x+1 = x+2-1, I do (x+1)+(1-1), so that I make sure I am not changing the equation by mistake. It comes in handy when dealing with bigger equations. Anyways, nice video!
my first course of action was to do integration by parts. after integrating by parts twice and doing a little bit of algebra i end up with a different answer (wrong): -ln|x+2| - x + C
Another trick with integration by parts, related to this idea, is that at each integration step, you can add any constant you want, since all we need is AN integral of the previous entry in the integration column, and any valid integral will work. We can add any arbitrary constant we want, and substitute the intermediate arbitrary constant that has an advantage. Most of the time, we keep it simple and just add zero. Example: integral x*ln(x^2 + 6) dx S _ _ D _ _ _ _ _ _ _ _ _ _ _ I + _ _ ln(x^2 + 6) _ _ _ _ _ x - _ _ 2*x/(x^2 + 6) _ _ _ 1/2*x^2 + B Construct IBP result: (1/2*x^2 + B)*ln(x^2 + 6) - integral (x^3 + 2*B*x)/(x^2 + 6) dx Factor numerator in integral: (1/2*x^2 + B)*ln(x^2 + 6) - integral x*(x^2 + 2*B)/(x^2 + 6) dx Wouldn't it be nice if (x^2 + 2*B) equaled (x^2 + 6)? It sure would, since that would cancel that part of the term. Let B=3 to make this so. (1/2*x^2 + 3)*ln(x^2 + 6) - integral x*(x^2 + 6)/(x^2 + 6) dx (1/2*x^2 + 3)*ln(x^2 + 6) - integral x dx Carry out final integral, add +C and we're done: (1/2*x^2 + 3)*ln(x^2 + 6) - 1/2*x^2 + C
very good trick! luckily i was taught that in school. The best way to do math is by getting to the solution in the easiest way possible and i think u sub or integration by parts would take too long.
My first idea is to both add and subtract 1 to the numerator, to just deal with "1-1/(x+2)", as it integrates immediately, with first term just giving us 'x' and the other one '-ln|x+2|'. We end up with x-ln|x+2|+C as an answer, but I will be honest that even though I know this trick, I started to think how to solve it only after being suggested that the trick does exist. Just like in chess! If you know that there is some tactic in the position and one correct move, it's so much easier to find it! 😀
It's a combination of observation, common sense & practice. At a glance it's obvious that X+2=(X+1)+1, so by numerator separation & division, the next step of a 'u' sub is apparent. However, if the denom was say X^2+1 (raised power eg X^n), then some more manipulation would be necessary. Inspection & practice, those are the key things with Integrals & yes, this one was WAY too easy!! 🤔
Polynomial long division is something you do for like one week out of your life and then nobody ever does it or makes you do it again for years, I assume either out of respect for students' or graders' time or otherwise not to muddle whatever the next lesson is; you pick up pretty quickly that you should not default to it. I would have done this the highlighted way, and if was asked for a second way, I would start fooling with Feynman's trick or something.
interesting perspective. Every student's cirriculum is different and many students are finding value from the video. Plus, for this simple problem, Feynnman's trick would be overkill anyhow.
Whats different here? thats how everybody solves this problem atleast in india when he said polynomial division i was like why tf would anybody do that such a simple step is to add 1 and subtract to get x+2-1 seriously i didnt knew that other countries use complex methods to solve such a simple problem
I solve it the way you do as well. But someone might think synthetic division is the way to go since the numerator polynomial degree isn’t less than the denominator
@@NumberNinjaDave It stands for Joint Entrance Exam (JEE). It is an entrance exam for engineering colleges in India and is considered one of the hardest exams in the world
I am getting it wrong for some reason! I got x + 2 - ln(abs(x + 2)) + c My steps Let u = x + 2 Integral becomes Integral((u - 1) / u, du) = Integral(u/u - 1/u, du) = Integral(1 - 1/u, du) = u - ln(abs(u)) + c = x + 2 - ln(abs(x + 2)) + c
If you take the derivative of (x+2) - ln|x+2| and the derivative of x - ln|x+2|, you will get the same function since the derivative of 2 is zero. 2 + c is another constant, so they are both antiderivatives of the integrand.
@@nightytime this is correct in terms of taking. the derivative still giving the sample problem. Remember, the indefinite integral gives you a *family* of functions of a generalized form with a constant C. You happened to find one of the family functions. But the answer isn't fully precise since we have an indefinite integral here.While the two functions indeed have the same derivative, the reverse direction must take into account a generalized form where an integral gives a 1 to infinitely many parent functions, also distinguished by the plus C constant I believe it comes down to your order of operations in separating out the integrals. Here's how I did it: let u = x+ 2 Then the (x+1) in the numerator of the original problem needs to be rewritten in terms of u, like you did. So since u = x + 2, I want the right to look like x + 1 and so I subtract both sides of the equation by 1, giving: u - 1 = x +1 Notice that this is a substitution so to be careful, I like to put in parantheses for order to be careful and deliberate on the order of evaluation: Integral ( (u - 1) / u ) du As you did so, separate this out into a difference of two fractions, with each one being its own integral problem Integral(u/u) du - Integral(1/u) du The first one simplifies to the integral of du, but remember that du = dx! So the integral of dx with respect to x is just x The right had one because ln | x +2|, giving the final answer that matches *if* if you add the + C at the end Note also in your answer, the extra term you had didn't take into account that your final answer has a + C anyhow so you can rewrite the sum of the C and your residual constant as a constant K if you want, to ensure your answer is a family of functions and not just one specific parent function. Hope that helps!
@@NumberNinjaDave Right, I was more so implying that the answer @thefreeze6023 got isn't necessarily incorrect. x + 2 - ln(abs(x + 2)) + c₁ can be rewritten as x - ln(abs(x+2)) + c₂, where c₂ = 2 + c₁.
@@nightytime yeah, I knew where you were coming from. My response was intended to buffer your response and clarify for him. I could have done a better job at that
Why use u-sub all the time even on easy integrals? You notice 1/(x+a) is a derivative/function so the integral of that is ln|x+a|. Simply use reverse chain rule if it's 1/(ax+b) and get ln|ax+b|/a
@@NumberNinjaDave well if you're clever enough to use u-sub on the denominator then you might as well guess the integral. I get that it's not obvious for everyone, but relying on u-sub too much ain't gonna do you any good
I did a u substitution
Let u = x + 2
u - 1 = x + 1 and dx = du
After doing that, getting to x - ln|x+2| + c is easy but your way is fascinating.
Very cool!
Same here
Same thing
I took u= x+2 and u-2=x but still got the answer
I usually do this method only but just write it out differently because I find that way more intuitive. Instead of directly write x+1 = x+2-1, I do (x+1)+(1-1), so that I make sure I am not changing the equation by mistake. It comes in handy when dealing with bigger equations. Anyways, nice video!
Thank you! Hey, that strategy is sound 👌
I think that's how everyone solves these anyways
Oh really
yeah same
Yeah everyone does this the only dumb people' can't get this
nah we doing long division
Same hear - add and subtract way. It is how schools and universities teach us.
When integrating, if integrating the denominator would lead to the same value as the numerator, it's going to lead to ln of the denominator
What about for polynomial degrees greater than 1
my first course of action was to do integration by parts. after integrating by parts twice and doing a little bit of algebra i end up with a different answer (wrong):
-ln|x+2| - x + C
Keep at it!
Another trick with integration by parts, related to this idea, is that at each integration step, you can add any constant you want, since all we need is AN integral of the previous entry in the integration column, and any valid integral will work. We can add any arbitrary constant we want, and substitute the intermediate arbitrary constant that has an advantage. Most of the time, we keep it simple and just add zero.
Example:
integral x*ln(x^2 + 6) dx
S _ _ D _ _ _ _ _ _ _ _ _ _ _ I
+ _ _ ln(x^2 + 6) _ _ _ _ _ x
- _ _ 2*x/(x^2 + 6) _ _ _ 1/2*x^2 + B
Construct IBP result:
(1/2*x^2 + B)*ln(x^2 + 6) - integral (x^3 + 2*B*x)/(x^2 + 6) dx
Factor numerator in integral:
(1/2*x^2 + B)*ln(x^2 + 6) - integral x*(x^2 + 2*B)/(x^2 + 6) dx
Wouldn't it be nice if (x^2 + 2*B) equaled (x^2 + 6)? It sure would, since that would cancel that part of the term. Let B=3 to make this so.
(1/2*x^2 + 3)*ln(x^2 + 6) - integral x*(x^2 + 6)/(x^2 + 6) dx
(1/2*x^2 + 3)*ln(x^2 + 6) - integral x dx
Carry out final integral, add +C and we're done:
(1/2*x^2 + 3)*ln(x^2 + 6) - 1/2*x^2 + C
@@carultch I love your thinking
very good trick! luckily i was taught that in school. The best way to do math is by getting to the solution in the easiest way possible and i think u sub or integration by parts would take too long.
Very true
My first idea is to both add and subtract 1 to the numerator, to just deal with "1-1/(x+2)", as it integrates immediately, with first term just giving us 'x' and the other one '-ln|x+2|'.
We end up with x-ln|x+2|+C as an answer, but I will be honest that even though I know this trick, I started to think how to solve it only after being suggested that the trick does exist.
Just like in chess! If you know that there is some tactic in the position and one correct move, it's so much easier to find it! 😀
nice approach
0 and 1 my favorite numbers. Add 0 or multiple by 1.
@@nathanluca3072 facts
wtf why your first thought is to long divide
my first thought was to do the thing shown
Good job.
It's a combination of observation, common sense & practice.
At a glance it's obvious that X+2=(X+1)+1, so by numerator separation & division, the next step of a 'u' sub is apparent.
However, if the denom was say X^2+1 (raised power eg X^n), then some more manipulation would be necessary.
Inspection & practice, those are the key things with Integrals & yes, this one was WAY too easy!! 🤔
What’s obvious to you may not be obvious to others. Otherwise, they wouldn’t be going to TH-cam for additional math help.
I instinctively think of quotient rule
to be fair the polynomial division here is quite short. but this is a nice technique for fractions with more terms
Yup
add and subtract one answer x-ln|x|
@@MathsandCoding you’re a ninja 🥷
is this called a trick in america? lol in india this is the only approach, other popular one is u sub but i rarely do it cus im lazy
Great! Everyone is different.
Use reverse chain rule instead of u sub
Reverse chain rule and u sub are analogous.
Polynomial long division is something you do for like one week out of your life and then nobody ever does it or makes you do it again for years, I assume either out of respect for students' or graders' time or otherwise not to muddle whatever the next lesson is; you pick up pretty quickly that you should not default to it. I would have done this the highlighted way, and if was asked for a second way, I would start fooling with Feynman's trick or something.
interesting perspective. Every student's cirriculum is different and many students are finding value from the video. Plus, for this simple problem, Feynnman's trick would be overkill anyhow.
makes sense
🙏
Whats different here? thats how everybody solves this problem atleast in india when he said polynomial division i was like why tf would anybody do that such a simple step is to add 1 and subtract to get x+2-1 seriously i didnt knew that other countries use complex methods to solve such a simple problem
I solve it the way you do as well. But someone might think synthetic division is the way to go since the numerator polynomial degree isn’t less than the denominator
Very easy for jee aspirants
What is that
@@NumberNinjaDave It stands for Joint Entrance Exam (JEE). It is an entrance exam for engineering colleges in India and is considered one of the hardest exams in the world
@@gametimewitharyan6665 oh, awesome. Thanks for explaining
@@NumberNinjaDave You are welcome :)
@@NumberNinjaDave make a video on JEE advanced questions you will get many views plzzzzzzz!
this is how i was originally taught (I'm indian)
That’s awesome 😎
🙏🙏
Viral this! Viral this on the internet!
That would be epic
That’s how everyone else does? No?
Not everyone
@@NumberNinjaDave so what is the actual method you all do?
@@Lucid.28 i do it the way shown in the video. I’ve seen people do polynomial division or slight modifications to what’s shown here
@@NumberNinjaDave ohh to divide it .. yeah if it’s more complicated , people would do it that way I think
This was short and to the point. 👍
Thank you
I would’ve done u substitution.
🎉
add and subtract one. --> x - ln|x| + c
I am getting it wrong for some reason! I got x + 2 - ln(abs(x + 2)) + c
My steps
Let u = x + 2
Integral becomes Integral((u - 1) / u, du)
= Integral(u/u - 1/u, du)
= Integral(1 - 1/u, du)
= u - ln(abs(u)) + c
= x + 2 - ln(abs(x + 2)) + c
If you take the derivative of (x+2) - ln|x+2| and the derivative of x - ln|x+2|, you will get the same function since the derivative of 2 is zero. 2 + c is another constant, so they are both antiderivatives of the integrand.
@@nightytime this is correct in terms of taking. the derivative still giving the sample problem. Remember, the indefinite integral gives you a *family* of functions of a generalized form with a constant C. You happened to find one of the family functions. But the answer isn't fully precise since we have an indefinite integral here.While the two functions indeed have the same derivative, the reverse direction must take into account a generalized form where an integral gives a 1 to infinitely many parent functions, also distinguished by the plus C constant
I believe it comes down to your order of operations in separating out the integrals. Here's how I did it:
let u = x+ 2
Then the (x+1) in the numerator of the original problem needs to be rewritten in terms of u, like you did. So since u = x + 2, I want the right to look like x + 1 and so I subtract both sides of the equation by 1, giving:
u - 1 = x +1
Notice that this is a substitution so to be careful, I like to put in parantheses for order to be careful and deliberate on the order of evaluation:
Integral ( (u - 1) / u ) du
As you did so, separate this out into a difference of two fractions, with each one being its own integral problem
Integral(u/u) du - Integral(1/u) du
The first one simplifies to the integral of du, but remember that du = dx! So the integral of dx with respect to x is just x
The right had one because ln | x +2|, giving the final answer that matches *if* if you add the + C at the end
Note also in your answer, the extra term you had didn't take into account that your final answer has a + C anyhow so you can rewrite the sum of the C and your residual constant as a constant K if you want, to ensure your answer is a family of functions and not just one specific parent function. Hope that helps!
@@NumberNinjaDave Right, I was more so implying that the answer @thefreeze6023 got isn't necessarily incorrect.
x + 2 - ln(abs(x + 2)) + c₁ can be rewritten as x - ln(abs(x+2)) + c₂, where c₂ = 2 + c₁.
@@nightytime yeah, I knew where you were coming from. My response was intended to buffer your response and clarify for him. I could have done a better job at that
X-ln|x+2|
Very close. It’s missing a small but important detail
@@NumberNinjaDave ha ha ,
you mean const. C ( C is going to Chill )
@@youngman3544 Yes sir 🥷
Why use u-sub all the time even on easy integrals? You notice 1/(x+a) is a derivative/function so the integral of that is ln|x+a|. Simply use reverse chain rule if it's 1/(ax+b) and get ln|ax+b|/a
@@erezsolomon3838 That’s not so obvious to every student. Feel free to use that method if you want
@@NumberNinjaDave well if you're clever enough to use u-sub on the denominator then you might as well guess the integral. I get that it's not obvious for everyone, but relying on u-sub too much ain't gonna do you any good
What psychopath would use polynomial division
Lol
I can write direct answer by doing these things in my mind because I m asian
@@kakashithecopyninja4426 hahaha right on! I do too but my videos are meant to help those who don’t see it
Ok... Then i will not judge you 😃
VERY COOL THANKS ALOT
Glad to help