Tricks | A QUICKER Way To Integrate This 🤯 

Very cool!

Same here

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.

Thank you! Hey, that strategy is sound 👌

Thank you

Very true

nice approach

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

🎉

What’s obvious to you may not be obvious to others. Otherwise, they wouldn’t be going to RU-vid for additional math help.

🙏

Yup

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

Glad to help

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

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.

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!

Reverse chain rule and u sub are analogous.

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 🥷

That’s awesome 😎

Great! Everyone is different.

Good job.

That would be epic

Not everyone

@@NumberNinjaDave so what is the actual method you all do?

@@user-er2nx5py7r 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

Lol