Stuck on an Integral: Need a Nudge in the Right Direction!

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Homework Help Overview

The discussion revolves around evaluating the integral \(\int \frac{(x+2)dx}{\sqrt{x^2+6x+4}}\), which involves techniques from calculus, specifically integration methods. The original poster expresses frustration with their attempts and seeks guidance on how to proceed.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts completing the square and substitution methods but finds their results unsatisfactory compared to an online integrator's output. Some participants suggest alternative substitutions, including hyperbolic functions.

Discussion Status

Participants are exploring various substitution methods and discussing the transformations of the integral. There is an exchange of ideas regarding the validity of different approaches, but no consensus has been reached on the correct solution.

Contextual Notes

There is mention of differing results from an online integrator and confusion regarding the manipulation of terms within the integral. The original poster's uncertainty about their answer highlights the complexity of the problem.

Zurtex
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Got an integral that for some reason I can't do, very much annoying me because this is supposed to be the stuff I am good at. A nudge in the right direction would be great, here it is:

[tex]\int \frac{(x+2)dx}{\sqrt{x^2+6x+4}}[/tex]

I have attempted completing the square on the bottom to get [itex](x+3)^2-5[/itex] and then I tried the substitution [itex]u=x+3[/itex]. After messing about with it a bit I got an answer but it didn't look right and it was a lot different from the answer I got off an online integrator and its answer really did look better.

Any help would be great thanks.
 
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Have you tried the substitution:
u= sqrt(5)*Cosh(t)?
 
Thanks, could someone please confirm this is the answer then:

[tex]\sqrt{ \frac{x^2-6x+4}{5} } - \cosh^{-1}\left(\frac{x-3}{\sqrt{5}}\right) + C[/tex]
 
Well, I got the -6x term to be +6x, and the x-3 to be x+3
 
Actually, I like your first idea of substituting for x+3. If you let u= x+3, then
x2+6x+ 4= (x+3)2- 5= u2- 5 and x+2= x+3-1= u-1 so the integral becomes
[tex]\int{\frac{u-1}{\sqrt{u^2-5}}du[/tex]
which we can separate into
[tex]\int{\frac{udu}{\sqrt{u^2-5}}+ \int{\frac{du}{\sqrt{u^2-5}}[/tex]

Now you can make the substitution v= u2-5 in the first integral and
[tex]sec(\theta)= \sqrt{5}u[/tex] in the second.
 
Last edited by a moderator:
kk thanks for the help and thanks for the other approach :smile:
 

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