You are Here: Home >< Maths

# Integration help watch

1. Hello, bit rusty on integration, but I have this integral: x^2/(x^2-2x+5) dx, and I've been told I have to use u-substitution. Tried doing u= x^2-2x+5 but haven't managed to get anywhere so far. Any tips?
2. (Original post by jordanwu)
Hello, bit rusty on integration, but I have this integral: x^2/(x^2-2x+5) dx, and I've been told I have to use u-substitution. Tried doing u= x^2-2x+5 but haven't managed to get anywhere so far. Any tips?
Complete the square on the denominator, THEN look for an appropriate substitution.
3. (Original post by jordanwu)
Hello, bit rusty on integration, but I have this integral: x^2/(x^2-2x+5) dx, and I've been told I have to use u-substitution. Tried doing u= x^2-2x+5 but haven't managed to get anywhere so far. Any tips?
Try doing long division (shouldn't be too difficult) first, then complete the square on the denominator as RDKGames suggested.
4. (Original post by RDKGames)
Complete the square on the denominator, THEN look for an appropriate substitution.
Ok so I have (u+1)^2 / u^2 + 4 du so far?
5. (Original post by jordanwu)
Ok so I have (u+1)^2 / u^2 + 4 du so far?
Expand the numerator, and say that to get on the numerator. Then split your fraction into 3 separate ones with numerators , and with each denominator being, yknow, unchanged. Then the first term simplifies to a constant, the second term is a matter of using the ln rule to integrate, and the third integrates to some trig function which you should recognise.
6. (Original post by RDKGames)
Expand the numerator, and say that to get on the numerator. Then split your fraction into 3 separate ones with numerators , and with each denominator being, yknow, unchanged. Then the first term simplifies to a constant, the second term is a matter of using the ln rule to integrate, and the third integrates to some trig function which you should recognise.
I ended up getting x-1 + ln((x-1)^2 + 4) - (3/2)tan^-1((x-1)/2) + C, not sure if I messed up somewhere or not
7. (Original post by jordanwu)
I ended up getting x-1 + ln((x-1)^2 + 4) - (3/2)tan^-1((x-1)/2) + C, not sure if I messed up somewhere or not
Looks good.
8. (Original post by jordanwu)
I ended up getting x-1 + ln((x-1)2 + 4) - (3/2)tan-1((x-1)/2) + C, not sure if I messed up somewhere or not
it looks correct
9. (Original post by RDKGames)
Looks good.
Ok, so the next one is this: 1/(5-2cosx) dx, and it says to use x= a tan^-1(u) for some appropriate constant a. Not sure where to start with this
10. (Original post by jordanwu)
Ok, so the next one is this: 1/(5-2cosx) dx, and it says to use x= a tan^-1(u) for some appropriate constant a. Not sure where to start with this
Why don't you use x=a arctan u and then choose an a halfway through the calculation when you see what is most useful?

### Related university courses

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: October 25, 2017
The home of Results and Clearing

### 2,905

people online now

### 1,567,000

students helped last year
Today on TSR

### University open days

1. Sheffield Hallam University
Tue, 21 Aug '18
2. Bournemouth University
Wed, 22 Aug '18
3. University of Buckingham
Thu, 23 Aug '18
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