# 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.

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