# Integration by Parts [Solved!]

**phinah** 11 Dec 2018, 14:54

### My question

When using IBP, is the rule to simplify v times du FIRST and then integrate that result, or is it to integrate 'v' first then multiply that result by du?

### Relevant page

7. Integration by Parts

### What I've done so far

Example 3 in this section:

After finding dv, if we integrate 'v' alone and then multiply the result by du, we end up with the answer `{x^3 ln 4x) /4` and not the answer on your web site attained by simplifying ‘v times du’ first, arriving at `(x^2)/3`, and then integrating that result.

But there was another exercise whereby we obtained the same results using both methods.

X

When using IBP, is the rule to simplify v times du FIRST and then integrate that result, or is it to integrate 'v' first then multiply that result by du?

Relevant page
<a href="https://www.intmath.com/methods-integration/7-integration-by-parts.php">7. Integration by Parts</a>
What I've done so far
Example 3 in this section:
After finding dv, if we integrate 'v' alone and then multiply the result by du, we end up with the answer `{x^3 ln 4x) /4` and not the answer on your web site attained by simplifying ‘v times du’ first, arriving at `(x^2)/3`, and then integrating that result.
But there was another exercise whereby we obtained the same results using both methods.

## Re: Integration by Parts

**Murray** 11 Dec 2018, 20:11

Phinah

I hope I understood your question correctly.

We can only ever integrate the items that come **before** the `dx` part. So this is OK:

`int (x^3)/3 dx/x = int \color{red}{\fbox{:(x^2)/3:}} dx` `=1/3 int x^2 dx = (x^3)/9 + K`

But this is NOT OK:

`int \color{red}{\fbox{:(x^3)/3:}} dx/x` `= 1/\color{red}{\fbox{:3:}} int \color{red}{\fbox{:x^3:}} dx/x` `=1/3 (x^4)/4 xx 1/x = (x^3)/12 + K`

It will nearly always result in a different (and incorrect) answer.

Can you point out an example where it appears it was done the second way?

X

Phinah
I hope I understood your question correctly.
We can only ever integrate the items that come <b>before</b> the `dx` part. So this is OK:
`int (x^3)/3 dx/x = int \color{red}{\fbox{:(x^2)/3:}} dx` `=1/3 int x^2 dx = (x^3)/9 + K`
But this is NOT OK:
`int \color{red}{\fbox{:(x^3)/3:}} dx/x` `= 1/\color{red}{\fbox{:3:}} int \color{red}{\fbox{:x^3:}} dx/x` `=1/3 (x^4)/4 xx 1/x = (x^3)/12 + K`
It will nearly always result in a different (and incorrect) answer.
Can you point out an example where it appears it was done the second way?

## Re: Integration by Parts

**phinah** 15 Mar 2019, 08:54

Okay I see how it is now done. Thank you.

X

Okay I see how it is now done. Thank you.

You need to be logged in to reply.