Helpppp

For this q why is v equal to x ??

The part i circled in red

For this q why is v equal to x ??

The part i circled in red

(edited 1 year ago)

Original post by Alevelhelp.1

Helpppp

For this q why is v equal to x ??

The part i circled in red

For this q why is v equal to x ??

The part i circled in red

It comes from solving dv/dx = 1

Original post by Alevelhelp.1

Helpppp

For this q why is v equal to x ??

The part i circled in red

For this q why is v equal to x ??

The part i circled in red

So Integration by Parts can be used when you can split what is to be integrated into a "v" part and a "du/dx" part. It's useful when you can split into one which is "easy" to differentiate. In this case v=x is easy to differentiate right? dv/dx=1 !!!

So all you need then is the other part, the bit which you want to integrate.

So you have arbitrarily chosen v=x to "work" using ingtegation by parts. You've chosen the v so as to be nice to differentiate.

In general integration by parts is like a riddle. You need to trial and error sometimes to get the two bits. If you can find something which differentiates to a constant, as in this case, then all you need is the bit which integrates. Combine the two using the formula for integration by parts. It's trickier as there is no "standard" procedure and you could chose many "v" and "du/dx" all getting back to the same answer. Similarly, sometimes there is no easy answer. Dont forget that the integration by parts formula includes an integration term in itelf, which is why getting a constant in the forumla is "nice".

int(v*du)=uv-int(u*dv)

The dv I have highlighted is ideally a constant, in your example 1. This makes the int(u*dv) nicer. Notice that the formula includes both the v and dv terms, but you are integrating over the dv term. And integration over a contant is much much easier so try and get a v which differentiates to a constant.

If you cant find a "nice" v which differentiates to a constant then you can get into an infinite loop quickly in that your answer includes another integral which may need integration by parts. Hope you follow? So look for a constant in the dv term

(edited 1 year ago)

Original post by Zürich

So Integration by Parts can be used when you can split what is to be integrated into a "v" part and a "du/dx" part. It's useful when you can split into one which is "easy" to differentiate. In this case v=x is easy to differentiate right? dv/dx=1 !!!

So all you need then is the other part, the bit which you want to integrate.

So you have arbitrarily chosen v=x to "work" using ingtegation by parts. You've chosen the v so as to be nice to differentiate.

In general integration by parts is like a riddle. You need to trial and error sometimes to get the two bits. If you can find something which differentiates to a constant, as in this case, then all you need is the bit which integrates. Combine the two using the formula for integration by parts. It's trickier as there is no "standard" procedure and you could chose many "v" and "du/dx" all getting back to the same answer. Similarly, sometimes there is no easy answer. Dont forget that the integration by parts formula includes an integration term in itelf, which is why getting a constant in the forumla is "nice".

int(v*du)=uv-int(u*dv)

The dv I have highlighted is ideally a constant, in your example 1. This makes the int(u*dv) nicer. Notice that the formula includes both the v and dv terms, but you are integrating over the dv term. And integration over a contant is much much easier so try and get a v which differentiates to a constant.

If you cant find a "nice" v which differentiates to a constant then you can get into an infinite loop quickly in that your answer includes another integral which may need integration by parts. Hope you follow? So look for a constant in the dv term

So all you need then is the other part, the bit which you want to integrate.

So you have arbitrarily chosen v=x to "work" using ingtegation by parts. You've chosen the v so as to be nice to differentiate.

In general integration by parts is like a riddle. You need to trial and error sometimes to get the two bits. If you can find something which differentiates to a constant, as in this case, then all you need is the bit which integrates. Combine the two using the formula for integration by parts. It's trickier as there is no "standard" procedure and you could chose many "v" and "du/dx" all getting back to the same answer. Similarly, sometimes there is no easy answer. Dont forget that the integration by parts formula includes an integration term in itelf, which is why getting a constant in the forumla is "nice".

int(v*du)=uv-int(u*dv)

The dv I have highlighted is ideally a constant, in your example 1. This makes the int(u*dv) nicer. Notice that the formula includes both the v and dv terms, but you are integrating over the dv term. And integration over a contant is much much easier so try and get a v which differentiates to a constant.

If you cant find a "nice" v which differentiates to a constant then you can get into an infinite loop quickly in that your answer includes another integral which may need integration by parts. Hope you follow? So look for a constant in the dv term

I know how to do by parts but I don’t get where the x came from because we are integrating (lnx)^2

so isn’t there lnx and ANother lnx??

Original post by Alevelhelp.1

I know how to do by parts but I don’t get where the x came from because we are integrating (lnx)^2

so isn’t there lnx and ANother lnx??

so isn’t there lnx and ANother lnx??

So are you suggesting doing IBP with u=ln(x) and dv=ln(x)dx? (standard IBP procedure btw)

Technically you can try, but (i) you need to integrate ln(x); and (ii) I have no idea if it will turn out to be useful or not. Most probably not useful though, I'd reckon.

(edited 1 year ago)

Original post by Alevelhelp.1

I know how to do by parts but I don’t get where the x came from because we are integrating (lnx)^2

so isn’t there lnx and ANother lnx??

so isn’t there lnx and ANother lnx??

No, (lnx)^2 is your u, so you are saying effectively "what would I need to integrate wrt x, to get back to u" or "what is my du/dx?"

if u =(lnx)^2 then du/dx=2*(lnx)/x. Therefore integral of (2*lnx)/x wrt x is (lnx)^2 (this is just an application of the chain rule basically)

You have a nice v, dv/dx=1

So in short you now have u, v, du/dx and dv/dx. You have everything you need to go into the formula for integration by parts.

Remember with integration by parts, look for a "v" which differentiates to a constant. Then you need a du/dx which INTEGRATES back to u nicely.

then you have a problem which is easily seperated out into 1) what differentiates and 2) what integrates and combine the two!

(edited 1 year ago)

Original post by Alevelhelp.1

Helpppp

For this q why is v equal to x ??

The part i circled in red

For this q why is v equal to x ??

The part i circled in red

If you try to split your product into (in x)(ln x) then you've got the problem of integrating ln x, which is possible but not obvious, and may lead to a more complicated integral after the 1st iteration of IBP.

Think about how you integrate ln x as a simpler example. You write it as 1 * ln x and then apply IBP, This is the same principle.

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