 Now we are going to start with the integration of my parts. So it is simpler than this. Normally what happens is no, in integration question whenever there will be p times this plus q times this, what is p? Integrate everything but you should know that in my answer I just have to find the coefficient of that term. So let's find it out, solve then. You don't have to complete the process. Depends upon the question asked. It is very real to get the answer. So before we start integration by parts, important especially with respect to school exams. This process, so both the types the approach is the same, just listen to the approach. Let me take an example, x to the power 4 minus, let's say, right now they have 1 plus 1 by x square is differentiation of which term? 1 plus 1 by x square is a differentiation of which term? x plus 1 by x. x minus 1 by x. So take x minus 1 by x as t, so that we get 1 plus 1. So can I say it will be square plus 1 by x square minus 2 is equal to t square. So x square plus 1 by x is equal to t square plus 2. So this term here you can replace it with t square plus 2. So we will get dt by t square minus 1 which is the formula of x square minus a square correct? x square minus t by x square. You divided by x square first. Yeah. Yeah, what is the answer to this? Shweta, what is the answer for this? x square minus 1 by x. x to the power 2 plus 1. x to the power 2 plus 1 by x. So put back your t as x minus 1 by x minus 1 by x minus 1 by x plus 1. You may take the x and same on both sides, so x square minus x minus 1 by x square plus x minus 1. Is the idea clear? This is the very very commonly asked time in your school exam, so be very very with the concept with the algorithm. If it helps take it, that is why I told try it out. If you take x square as t you should have an x dx on the top which is not available. Now here one quick thing x is t, no psi plus 1. How do I integrate this? Please allow me to clear this. Rishali copies. For such minus x square minus 1. Then we separate these x square plus 1. And you have to do it both separately each one of them. So you know how to do this. Cut by x square take x minus 1 by x s t. Divide by x square take x plus 1 by x s k. Two forms, both the manipulations I have told you. I am making you manipulative. One favorite question that is asked in this time is integral or I got this for my body exams. What you are going to do? Second and third question you definitely remember it for the rest of the life. You can solve this one. Yeah but after a lot of solving. This is this side yeah. See you have to write tan it is one square. So think how it is of this type and then how do you solve this? Answer. Sir you have to write this. Sir but then how do you write this? Then how will we know that this type is written like this. But second sign is then you come to a form of square plus 1 and x square minus 1. I know usually it is not psychic. But once you see this at least that idea will be there in a moment. I can't do that. Hit. Hit. Hit. Hit is everything. Sir you have to write tan. Multiply and divide with c square x and then substitute tan x as c square x. See what happens. Multiply and divide with c square x and then put tan x as c square x. See it. Multiply and divide with. This part is fine. Write the c square as 1 plus tan square. And then I ask you to take tan x as c square. So that c square x dx is 2t dt. Now when you substitute here. Part 4. 2t square is this what you are getting finally. Yeah. Yes. So this 2t square by God's grace can be written like this. 4. There is 4 months. Yes. So this column is stood between me getting an undead and 96. Okay. So then divide by t square. So you get 1 plus 1 by t square by t square plus 1 by t square. Sir how long do we spend on this question? 15 minutes. You would realize when you start getting in for use the integration problems. They will kill. They are known to kill a lot of people. What do you substitute as? Let's call it as p. Which implies 1 plus 1 by t square dt will be equal to dp, right? p square plus 2. p plus 1 by t first. So it will be 1 by root 2 minus 1 by root tan x. And in the second expression you will have minus root 2 will be k is root tan minus root 2 by root tan x plus 1 by root tan x plus root t. It is like the c. It is simple. Yeah. So imagine the fate of that whole child. I think one marks a loss in probability. Root tan x plus root cortex. That is simpler than this actually. Even though they have more terms. This is more difficult than solving root tan x plus root cortex. Would you like to try? Yes. Integrate root tan x plus root cortex. Again a very famous problem you may come in your school exams as well. Let's see about which. Exactly the same one. The question is x minus 1 by x as t and where to take x plus 1 by x as t. You are numerator.