# ∫(4x - 3) / (x^2 +1) dx

1. Jun 1, 2010

### jasper10

1. The problem statement, all variables and given/known data

Can someone help me integrate (4x - 3) / (x^2 +1) ?

3. The attempt at a solution

I'm doing my maths baccalaureate in 2 days and came across this question today!

I don't know where to start i.e. which method to use:

Integration by parts definitely does not work, nor does by substitution!

Thank you!

2. Jun 1, 2010

### tiny-tim

Hi jasper10!

Trig subsitution should do it.

(alternatively, split it into the 4x part and the 3 part … you should be able to integrate those anyway )

3. Jun 1, 2010

### DJsTeLF

easy to do if you use integration by parts since the derivative of the numerator is a constant :D

4. Jun 1, 2010

### Staff: Mentor

I think that tiny-tim's advice of splitting the integral into two parts is much simpler than integration by parts - hence, less opportunity for getting things fouled up.

5. Jun 1, 2010

### DJsTeLF

Unless I'm sorely mistaken integration by parts is one of the key things the maths part of an IB is supposed to teach you. I think my hint was sufficient and that if appropriate care is taken nothing should get 'fouled-up'.

6. Jun 1, 2010

### jasper10

Ok so i get:

∫4x / (x^2 + 1) dx - ∫3/ (x^2 + 1) dx

= 2ln(x^2 + 1) - ???

How would you integrate: 3 / (x^2 + 1)?

the annoying part is the "x^2"

thanks!

ps: what do you mean by "trig" substitution?

(I have a feeling this type of question isn't on the syllabus, but i found it on a past back from 2002 - strange)

7. Jun 1, 2010

### jasper10

Integration by substitution surely cannot work!

it's a vicious circle and keeps getting more complicated whilst integrating

8. Jun 1, 2010

### Staff: Mentor

I'm not disputing the importance of integration by parts. What I'm saying is that it's always a good idea to use the simplest technique that seems likely to work. For this problem, integration by parts ain't it.

9. Jun 1, 2010

### DJsTeLF

Yeah just had another look at it and realised the complication, my apologies.

To quote myself, apparently I was "sorely mistaken" :S

10. Jun 1, 2010

### Staff: Mentor

Apparently you mean "integration by parts."

11. Jun 1, 2010

### jasper10

I don't know, nothing seems to work

I'm off to bed now, maybe tomorrow i can think more clearly.

12. Jun 1, 2010

### Staff: Mentor

$$\int \frac{dx}{1 + x^2} = tan^{-1}x + C$$

Presumably you've run across this antiderivative formula and variations of it.

13. Jun 2, 2010

### tiny-tim

good morning!
"trig" substitution is the unofficial but widespread abbreviation for substitution with a trigonometric or hypertrigonometric function (in this case, x = tanu or x = sinhu)

but you should have known this one anyway, for the exam …

check the PF list of standard integrals and memorise a few

14. Jun 2, 2010

### Susanne217

European baccalaureate or American baccalaureate??? Never heard about the term before ?

15. Jun 2, 2010

### jasper10

European baccalaureate..why? :D

16. Jun 2, 2010

### jasper10

Ok, thanks Mark44 and tiny-tim!

17. Jun 2, 2010