Need help integrating (4x - 3) / (x^2 +1) for my Maths Baccalaureate exam!

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Homework Help Overview

The original poster seeks assistance with integrating the function (4x - 3) / (x^2 + 1) in preparation for their Maths Baccalaureate exam. The problem involves techniques of integration, particularly focusing on methods such as integration by parts and substitution.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss various methods for integration, including integration by parts and trigonometric substitution. Some suggest splitting the integral into simpler parts, while others express uncertainty about the effectiveness of these methods. The original poster questions the applicability of integration techniques and expresses frustration with the complexity of the problem.

Discussion Status

There is an ongoing exploration of different integration techniques, with participants providing hints and suggestions. Some express doubts about the appropriateness of certain methods, while others emphasize the importance of using simpler techniques. The discussion reflects a range of interpretations and approaches without reaching a consensus.

Contextual Notes

The original poster mentions a time constraint due to an upcoming exam and expresses concern about the relevance of the problem to their syllabus. There is also a reference to a past exam question, indicating potential uncertainty about the material covered.

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Homework Statement



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

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!
 
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Hi jasper10! :smile:

Trig subsitution should do it.

(alternatively, split it into the 4x part and the 3 part … you should be able to integrate those anyway :wink:)
 
easy to do if you use integration by parts since the derivative of the numerator is a constant :D
 
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.
 
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'.
 
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)
 
DJsTeLF said:
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'.

Integration by substitution surely cannot work!

it's a vicious circle and keeps getting more complicated whilst integrating
 
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.
 
Mark44 said:
For this problem, integration by parts ain't it.

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

To quote myself, apparently I was "sorely mistaken" :S
 
  • #10
jasper10 said:
Integration by substitution surely cannot work!

it's a vicious circle and keeps getting more complicated whilst integrating
Apparently you mean "integration by parts."
 
  • #11
I don't know, nothing seems to work

I'm off to bed now, maybe tomorrow i can think more clearly.
 
  • #12
jasper10 said:
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)

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

Presumably you've run across this antiderivative formula and variations of it.
 
  • #13
good morning! :smile:
jasper10 said:
I'm doing my maths baccalaureate in 2 days and came across this question today! …
jasper10 said:
How would you integrate: 3 / (x^2 + 1)?

the annoying part is the "x^2"

thanks!

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

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

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

check the PF list of standard integrals and memorise a few :wink:

good luck in your exams! :smile:​
 
  • #14
tiny-tim said:
good morning! :smile:



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

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

check the PF list of standard integrals and memorise a few :wink:

good luck in your exams! :smile:​


European baccalaureate or American baccalaureate? Never heard about the term before ?
 
  • #15
European baccalaureate..why? :D
 
  • #16
Ok, thanks Mark44 and tiny-tim!
 
  • #17
jasper10 said:
European baccalaureate..why? :D

Hadn't heard the term before so I googled it. You can check your abilities to integrate

with the Wolfram integrator!
 

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