Difficult indefinite integral (mix of integration by parts and/or substitution)

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  • #1
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Homework Statement



I do not know how to solve the following indefinite integral.
I personally think it is very difficult and would appreciate it had
someone can explain it step by step?

integraal_1.png


Homework Equations



/

The Attempt at a Solution



This integral must been solved by mix of integration by parts and/or substitution
but I had no luck to get the answer.

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
SteamKing
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Hint: look at the numerator (without the ln(x) term) and look at the denominator.
Doesn't it look like a u and du perhaps?
Have you tried any substitutions and/or integrated by parts yet?
 
  • #3
haruspex
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The substitution w = x+1 might make SteamKing's hint easier to follow.
 
  • #4
SteamKing
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The substitution w = x+1 is really not helpful here, IMO.
However, if you take the derivative of the denominator, and then compare it to (x+1), you are getting warm.
 
  • #7
Curious3141
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You forgot to cube the denominator.
I can't see a cube on the denominator in the image that the OP posted. Maybe it's something to do with the way the image is rendered in my browser (Chrome). :confused: (Yes, I tried zooming, etc.)

I even tried using Firefox, to no avail.

What I'm seeing is:

[tex]\frac{(x+1)\ln(x)}{(x^2 + 2x + 10)}[/tex]

I don't even see the integral sign, element of integration, etc. (if they're there in the original image).

With the cube, it becomes easy.
 
  • #8
Dick
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I can't see a cube on the denominator in the image that the OP posted. Maybe it's something to do with the way the image is rendered in my browser. :confused: (Yes, I tried zooming, etc.)

I even tried using Firefox, to no avail.

What I'm seeing is:

[tex]\frac{(x+1)\ln(x)}{(x^2 + 2x + 10)}[/tex]

I don't even see the integral sign, element of integration, etc. (if they're there in the original image).

With the cube, it becomes easy.
Yeah. Your image is getting cut-off. I'm using Firefox on Ubuntu and it looks fine all the way to the dx to me.
 
  • #9
HallsofIvy
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My first thought was to let [itex]u= x^2+ 2x+ 10[/itex], the denominator, since its derivative is [itex]2x+ 2= 2(x+ 1)[/itex], but, of course, the "ln(x)" messes that up.

Okay so my second thought was to let [itex]u= ln(x)[/itex] so that [itex]du= (1/x)dx[/itex], [itex]xdu= e^u du= dx[/itex]. Since [itex]x= e^u[/itex], the integral becomes
[tex]\int \frac{e^u+1}{e^{2u}+ 2e^u+ 1}e^udu= \int \frac{e^2u+ e^u}{e^{2u}+ 2e^u+ 1}du[/tex]
and now try the substitution [itex]v= e^{2u}+ 2e^{u}+ 1[/itex].
 
  • #10
SteamKing
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The problem in the OP has the denominator cubed, and the constant term in the denominator is 10 and not 1.

The substitution x = e^u transforms the numerator to u* (Exp(u) + 1), and of course dx= Exp(u) du. Therefore, the numerator becomes:
(Exp(2u) + Exp(u))*u du.
 
  • #11
SammyS
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I was waiting to see if OP would return. That's looking doubtful.

It seems to me that a substitution of w = x+1, followed by integration by parts is the way to go. This substitution does look helpful. (I've tried it both ways.) After this it appears that partial fraction decomposition is in order. WolframAlpha shows a result with 5 terms which is consistent with all of this.
 
  • #12
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Homework Statement



I do not know how to solve the following indefinite integral.
I personally think it is very difficult and would appreciate it had
someone can explain it step by step?

integraal_1.png


Homework Equations



/

The Attempt at a Solution



This integral must been solved by mix of integration by parts and/or substitution
but I had no luck to get the answer.
Didn't the OP say we needed integration by parts?

[itex]\int \frac{(x+1)ln(x)}{(x^2 + 2x + 10)^3} \ dx= \int ln(x) \frac{(x+1)}{(x^2 + 2x + 10)^3} \ dx[/itex]. Let u = ln(x) (implying du = dx/x) and dv = (x+1)/((x^2 + 2x + 10)^3) dx. Thus, [itex]v = \int \frac{(x+1)}{(x^2 + 2x + 10)^3} dx= \frac{-1}{4(x^2 + 2x + 10)^2}[/itex]

[itex]\int ln(x) \frac{(x+1)}{(x^2 + 2x + 10)^3} \ dx = \frac{-ln(x)}{4(x^2 + 2x + 10)^2} - \int \frac{-1}{4(x^2 + 2x + 10)^2} \frac{dx}{x} \\
\frac{-ln(x)}{4(x^2 + 2x + 10)^2} - \int \frac{-1}{4x(x^2 + 2x + 10)^2} \ dx \rightarrow \frac{-ln(x)}{4(x^2 + 2x + 10)^2} - \frac{\frac{30(x-8)}{x^2 +2x + 10}+28 arctan\frac{(x+1)}{3}-54 ln(x)+27 ln(x^2 + 2x + 10)}{21600}[/itex]

I used the "\rightarrow" to indicate skipping the dreary use of partial fractions. I think you guys can take it from here...
 
  • #13
haruspex
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Didn't the OP say we needed integration by parts?
It said and/or.
I think you guys can take it from here...
With the possible exception of the OP poster, I think we already did. :wink:. (I've never been sure whether 'OP' stands for Original Post or Original Poster, but I've always taken it to be the former.)
 

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