You are Here: Home >< Maths

# When to use integration by parts or U-Substitution ? Watch

1. Just wondering is there a rule of thumb to know which technique to approach before attempting the question ?

For differentiation it was pretty straight forward to see the difference, but is there a difference for integration ?

Also, slight off topic but for partial fractions decomposition, is there a way to check if the solutions for the constants of the decomposition are correct ?

2. (Original post by T-GiuR)
Just wondering is there a rule of thumb to know which technique to approach before attempting the question ?

For differentiation it was pretty straight forward to see the difference, but is there a difference for integration ?

Also, slight off topic but for partial fractions decomposition, is there a way to check if the solutions for the constants of the decomposition are correct ?

For the partial fractions question, just sub a number in for x into the decomposition and the original fraction. If you get the same number from both of them, you've done it right
3. (Original post by Gome44)
For the partial fractions question, just sub a number in for x into the decomposition and the original fraction. If you get the same number from both of them, you've done it right
If I write , then when I put , I get the correct equation , but the decomposition is wrong so this method doesn't work in general.

To check the decomposition, you should recombine the final decomposition algebraically to see that you get back the original expression.
4. (Original post by T-GiuR)
Just wondering is there a rule of thumb to know which technique to approach before attempting the question ?

For differentiation it was pretty straight forward to see the difference, but is there a difference for integration ?

Also, slight off topic but for partial fractions decomposition, is there a way to check if the solutions for the constants of the decomposition are correct ?

Integration by parts = product rule. So whenver you have a product try to integrate by parts.
5. It's not as simple as 'being a product' as some products come from using the chain rule so they can be integrated without integration by parts.

For integration by parts you are looking for two functions which are not 'connected' like xsinx needs integration by parts but x(x^2 + 5)^3 doesn't.

There are also some functions you have to remember like ln x that you have to use parts by using 1. ln x
6. (Original post by atsruser)
If I write , then when I put , I get the correct equation , but the decomposition is wrong so this method doesn't work in general.

To check the decomposition, you should recombine the final decomposition algebraically to see that you get back the original expression.
Yes, but you could try a more obscure number, eg 2e^2
7. i think most of the time, knowing which choice is appropriate is just experience.

For sure, when you see a product (or quotient) of 2 functions where one is essentially the derivative of the other your choice is easy - sub.

Very often ones with roots in them (square or otherwise) are usually trig/hyperbolic trig subs.

As Muttley79 says, ones that appear not to be related are most likelly by parts.

But again, it comes down to the complexity of the function, as there are some functions that are more "suited" to one method and not another.

Example: what is the indefinite integral of

I use (especially in working out something called Fourier Series) a "cheats" way of IBP, called tabular integration - which, in fact you dont even need here, as practice will tell you the integral is multiplied by the alternating plus and minus successive derivatives of (including the "zero-ith" derivative)

viz:

Its worth knowing this instead of slogging through 5 or more pages of integration!

I dont recommend this to everyone, but, personally i dont care what the method is as long as my method is sound, and the answer is correct!
8. (Original post by atsruser)
If I write , then when I put , I get the correct equation , but the decomposition is wrong so this method doesn't work in general.

To check the decomposition, you should recombine the final decomposition algebraically to see that you get back the original expression.
The numerators of the decomposition would be constants so your example is not realistic
9. (Original post by T-GiuR)
Just wondering is there a rule of thumb to know which technique to approach before attempting the question ?

For differentiation it was pretty straight forward to see the difference, but is there a difference for integration ?

Also, slight off topic but for partial fractions decomposition, is there a way to check if the solutions for the constants of the decomposition are correct ?

In general they will give you the substitution and otherwise it will probably be by parts.To look for ones you should integrate by parts you need to think about what is going on you need two products one of which will reduce easily and eventually disappear as long as you have a reasonably easy function to integrate alongside it (eg. x^5e^x is integration by parts as x^5 will reduce by a power of 1 through each differentiation and eventually disappear and lnx is integration by parts as lnx will differentiate to something much more manageable 1/x) or form some expression that can be integrated by recognition(these can be harder to spot).Integration by substitution is usually used when there seems to be some portion of the function that is very troublesome that you want to substitute out and it is generally related to the rest of the function or trigonometric substitution is used when an expression looks similar to a trig identity like integrating sqrt(1-x^2).
10. Three cases for parts, in A-level maths at least:

- product of poly and exponential: e.g. xe^x
- product of poly and trig function: e.g. x sin x
- product of poly and log e.g. x ln x

For the first two, you usually differentiate the polynomial 'bit' and integrate the exponenital bit i.e. in the formula, u = x and dv/dx = e^x (or sin x...)

Third case you typically do it the other way round, so u = ln x and dv/dx = x in the above example.

TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: January 22, 2015
Today on TSR

### Should I ask for his number?

Discussions on TSR

• Latest
• ## See more of what you like on The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

• Poll
Useful resources

### Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

### How to use LaTex

Writing equations the easy way

### Study habits of A* students

Top tips from students who have already aced their exams

## Groups associated with this forum:

View associated groups
Discussions on TSR

• Latest
• ## See more of what you like on The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

• The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE