simple integration problem: which answer is correct of the two, "0.5" or "-2"?

[steveB]

I can't understand your question. You seem to mention what x should be and mention 0.5 and 1.5. However, the question is very clear about what values of x too use.

EDIT: I see now. You find the table in the solution manual seems inconsistent with the question. I would say answer the question and dont' worry about the solution manual.

 

[PG1995]

Hi

Could you please help me to solve this integral? Thank you.

Regards
PG

 

[steveB]

This is a good example where your "picture based" attachment is not being conducive to efficiently answering the question. I can't cut an paste from your work and easily point out the problem and I can't spare the time to rewrite everything in Latex or write it all out and scan it. But, no matter, I'll do my best to point to the issues.

First, there seems to be a typo in what you wrote because the denominator should have a u^2 in it, not a u.

Also, at the start of the third line, you had that factor of 2 factored out. It would be better if you not factor that back in. Instead keep the 2 where it is and then when you go back to the integral just factor a 1/2 constant in front of the integral.

I think if you pay attention to these two issues, you should be able to work through it.

 

[Ratchit]

Hi

Could you please help me to solve this integral? Thank you.

Regards
PG

Why don't you use http://integrals.wolfram.com/index.jsp?expr=(3x+1)/(2x^2-2x+3)&random=false ? I am sure it can do the problem a lot better and faster than any of us can. I am surprised you have not run across this program before. Sometimes it is more efficacious to look around a bit before you post.

Ratch

 

[PG1995]

Thank you, Steve.

I have solved it partially but can't figure out rest of the solution. Please help me. Thanks.

Regards
PG

 

[steveB]

PG,

Solving that remaining integral is easy if you use substitution for the variable u. Let w=cu, where c is chosen to make get the form K/(w^2+1), where K is a constant, as needed to make it work.

For example, if I had y=3/(u^2+7), i would multiply top and bottom by 1/7, and get the following.

y=(3/7)/(u^2/7+1)

Now, let w^2=u^2/7, and you will get the following.

y=(3/7) * (1/(w^2+1))

Or, some variation of thatapproach will work, as long as you arrive at the form 1/(w^2+1) that can be integrated more easily. Once you know the value of c, you can work out du and do it out.

 

[PG1995]

Thanks, Steve.

I'm sorry to ask you about this again. I have almost solved it but my final answer is still a bit different from the one given by my calculator. Could you please give it look? Thank you.

Regards
PG

 

[steveB]

PG,

I believe your answer is correct. The calculator answer has a different factor in front of the arctan function. By the way, you should give your answer as a mathematically exact form. Don't use approximate decimals. This means, keep the sqrt(5) 's in the answer. You know that already and just wanted to compare to the calculator. Still, if the calculator gives answers that way, use something else like Matlab or Wolfram.

Also, dont' forget that any indefinite integral should include an additive constant. We often leave it out, but it is important to remember it is there. For example, you were worried that the argument of the log function was off by a factor of 2. That's no big deal because any multiplication of a constant inside can be represented as an addition of the log of that constant outside.

That is ...

log (c x) + C1 = log(x) + log(c) +C1=log(x) + C2

Hence, either log(c x) or log(x) can be the indefinite integral of a function. If one works, then the other works.

 

[MrAl]

Hi,

For the original question it seems like the only thing missing was the squared sign for the 'u' in the denominator. The second line of equations moves from the full quadratic form to the 'u' form but the 'u' does not have a square as it should: 2*u instead of 2*u^2.
Thus it would not be possible to get the right result from that second line without including the square for the 'u'. Once the square is there it should probably work out ok following the results of the example given.

 

[PG1995]

Thank you, Steve, MrAl.

Here is the complete answer. You were right that it's the calculator answer which has a slightly different factor in front of arctan. And I myself converted the fractions into decimals because I was finding handling of fractions a bit difficult and it was taking a lot of space but I do understand your point.

Why don't you use http://integrals.wolfram.com/index.jsp?expr=(3x+1)/(2x^2-2x+3)&random=false ? I am sure it can do the problem a lot better and faster than any of us can. I am surprised you have not run across this program before. Sometimes it is more efficacious to look around a bit before you post.

Ratch: I was interested in the method and not in the final answer and I still have quoted the end solution in my main qustion. Anyway, thanks.

Regards
PG

 

[Ratchit]

PG,

I plotted the exact integral in red color and your approximation in blue. I also subtracted the constant 0.46 from the exact integral to bring both curves into alignment.

Ratch

 

[PG1995]

Thank you, Ratch.

I believe you needed to subtract 0.46 from the exact integral for the reason that I had used decimal representation instead of exact fractions.

Regards
PG

 

[steveB]

PG,

The actual constant should be 3*ln(2)/4 which is about equal to 0.5199. It's not the round off error, but the factor of 2 within the log function, as I explained before. Also, I don't think the round off error should produce as much error as shown in Ratch's plot. I would plot it yourself and make sure.

EDIT: Actually, I just did it myself and attached the plot. Note that the two curves match better than shown by Ratch's plot. I'm not sure what the issue is.

Still, the point is that it's best to use the exact answer when you can get it. But, the exact answer is only the exact answer within an additive constant.