# Integration by Parts [Solved!]

### 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

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

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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