Integration by part

[00:00:00]two function multiply each other. So the part by part if we have integral of fx* what g of x dx it's equivalent to what sheet* >> f * xus what >> integral of frime * what g of x so we can use this integration If we have two function multip and we said in order for us to identify f of x and g of x we need to identify leag so we said we'll be using the arrangement called le we said there are many arrangement that we can use so we said l stand for logarithmic function i stand for the inverse a stand for what? Right? T stand for trigger doain. E stand for exponential.

[00:01:08]C stand for a constant.

[00:01:13]So now we given equation there whereby we given what 1 / a b mult* by the what? Inverse of bx over what? A. So we want to integrate. So what we are seeing we are saying two function multiply each other. Why we say two function? Because we have our constant 1 / a it's a constant. So simply means if we supposed to use our arrangement the tech we can say we have this function the constant and we also have what this is called what inverse of trigonometry. So we have what? So we can say which one is first between the two according to the arrangement.

[00:02:03]>> Inverse. So we name inverse as one and then C as what two C. So the first function that we have identified there it tell us what function we have first.

[00:02:15]I mean the first function is f of x according to for so because we have identified one here. It means our f of x is what? It's inverse of three. So you're going to say what? A tan of what?

[00:02:28]b x / a. So the second function will be coming in 1.

[00:02:38]>> So therefore we find a derivative of this function. What is the derivative of this? We use the formula sheet.

[00:02:49]different x.

[00:02:57]So what is the derivative of f of x? Of course it will be what?

[00:03:01]>> B / a when you derive 3x what's the derivative of 3x? You said it's the constant.

[00:03:09]>> So what is the derivative? Our constant is b over a. So it means v / a. You put the derivative of the numerator and then here formulation to tell us it is 1 + f of x² and then our f of x is the whole function that we have b x over what a raised to power 2. So if you want you can apply the laws of exponent that this can be square to b and square to the what x² and also square to what to a. So this when we write it in a simple form we can say basically the derivative of that b what b² x what x² over what >> a square conant?

[00:03:57]>> Yeah. So here this side we integrate you when you integrate a constant you know this it will be 1 / what a b x we looking for the integration this side.

[00:04:11]So we are integrating this and then we do what so our job or our first step is to substitute it to the formula of part by part.

[00:04:25]Yeah. So therefore simply means since we know the formula part by part therefore we'll arrive and say we have our f of x what's our f of x we have already identified so it means of what bx / a multip by your g you also have your g which is 1 / what a b x minus Remember minus is part of the formula of part by partus.

[00:04:59]Yeah. Minus integral. What is derivative of fx? Is this one? So we're going to say a / b over 1 + b² x² a multiply by what?

[00:05:18]g of x. What is our g of x? You said it's 1 / a bx.

[00:05:25]Don't forget also to put the x because it's also there on our formula. So therefore we use our formula of part. So when we check the first two terms already does not have the integration sign. So this tell us this is already part of the solution or it's already our solution. So simply means we can leave it the way it is. Some would prefer to start with things the order does not matter because it's multiplication. So already it's part of our solution. But because here we have the integral side tell us that we must integrate.

[00:06:03]So number one we can start by simplifying this. So how can we simplify this is to identify the constant outside the integral.

[00:06:13]>> So what's our conant? Remember there on the numerator you have this >> and then on the other side you have what?

[00:06:25]>> 1 right.

[00:06:29]>> So what will happen?

[00:06:32]B and B cancel or if you want to write this like that this and this cancel right we so we take 1 a as our constant >> 1 a² a * a >> okay therefore we are left with what we'll be left with and then denominator you have the the way it is.

[00:07:18]So from here it simply means when you will have to integrate this.

[00:07:24]So you using your method or you use what your inspe is you know this will be what zero for one you going to inver So when you invert two it will be 2 b² a x. So when I will be able to cancel with x. So we'll make what dx will be the sub of the formula simply means it will be duid by the derative when we use it it was able to cancel the x term. So now when we substitute this in terms of don't forget we already have part of the solution - 1 / a². So when you come to this part you want to replace everything in terms of what?

[00:08:37]In terms of eu.

[00:08:40]So you have x. So a x will remain like that.

[00:08:48]So remember we already have part of the solution t to this it was already part of the solution we focus on the integral part so on the integral part remember now we want to replace in terms of eu. So x it's like that we say this it's already what our u and then dx we already know what is dx we replace also the 2b² over what a² x so what will happen you see x and x will do what will cancel so it simply means we can integrate that when x cancel what are we left with there remember this is part of the solution we'll be left with a constant So what we must do? We can identify the conant outside.

[00:09:42]>> So this is 1 / 2 b^ 2 a 2 simply means a 2 over what? 2 b². We have 1 over 2 b^ 2 over a 2. So mathematically this will be what? a 2 over what? 2 b². So we take out the constant. So we're going to take this use a multip 1 a. So we can simplify it.

[00:10:14]Outside we have 1 / a². We are taking out the constant of what? a² over what?

[00:10:20]2 b. So what will happen? A 2 cancel. So it means we'll be left with a conant of 1 / 2 b². We are three.

[00:10:33]>> Then lastly, one.

[00:10:36]What is one over u this is a conant. It remain the same.

[00:10:54]Then we open our brackets. What's the thing? You say this is U plus C. So at the end of the day, we're going to replace the value of U because we know what is the value of you.

[00:11:07]>> Yeah.

[00:11:09]>> Any question?