You are Here: Home >< Maths

# Integrals and divergence/convergence watch

1. Hi,

I have just done about three questions all working out whether the integrals converge or diverge:

∫dx/((x+4)(5x+1)) between infinity and zero
∫dx/x(1+x^2) between 1 and 0
∫dx/x^2(1+x^2) between 1 and 0

I have tried working them out by splitting them all into partial fractoins etc. but i am not sure if my answers are right, is there a quick way of checking my answers?

thanks
3. for the second and third ones i got that they both tended towards infinity as delta tends towards 0+ so they both diverge?

For the first one i got stuck
4. First one has an exact answer.
5. i got as far as -1/19[ln(R+4/5R) -ln4) im not sure if thats right?

Also do you know if the last two both tend to infinity and diverge?

Thanks
6. The second and third both diverge since you have an infinity of order greater than or equal to one in the denominator .
7. right thanks so im right that the last to tend to infinity and so diverge

but with the first one will i not just get that it diverges because does ln(R+4/5R) not tend to infinity as R tends to infinity?
8. (Original post by sonic7899)
right thanks so im right that the last to tend to infinity and so diverge

but with the first one will i not just get that it diverges because does ln(R+4/5R) not tend to infinity as R tends to infinity?
No it wouldn't.

Dodgy maths warning below before I get told off:

Try investigating it on your calculator using increasingly large values for R.

As R approaches infinity, you can effectively ignore the +4 bit.

What do you think the value of ln((R+4)/5R) might tend towards now?

If you are still not sure, break up the logarithm:

ln((R+4)/5R) = ln (R+4) - ln R - ln 5

Does that help?
9. yeh i think so, the whole thing will tend to -1/19[ln5-ln4] btw should what you but above not be (lnR + ln5)?
10. sorry to be a pain but can someone confirm this lol its driving me mad
11. anybody???
12. you should find it tends to 1/19[ln5 + ln4] i.e 1/19 ln20 check your workings for a sign error
13. thanks
14. i think the 1/19 has to be negative doesn't it because that comes from splitting the original expression in to partial fractions so surely -1/19[ln5-ln4] is correct?

### 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: November 17, 2008
Today on TSR

### Edexcel C4 Maths Unofficial Markscheme

Find out how you've done here

### 2,492

students online now

Exam discussions

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

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