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

1. Jan 8, 2013

### hoeranski

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

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?

2. Relevant equations

/

3. The attempt at a solution

This integral must been solved by mix of integration by parts and/or substitution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Jan 8, 2013

### SteamKing

Staff Emeritus
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. Jan 8, 2013

### haruspex

The substitution w = x+1 might make SteamKing's hint easier to follow.

4. Jan 9, 2013

### SteamKing

Staff Emeritus
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.

5. Jan 9, 2013

6. Jan 9, 2013

7. Jan 9, 2013

### Curious3141

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). (Yes, I tried zooming, etc.)

I even tried using Firefox, to no avail.

What I'm seeing is:

$$\frac{(x+1)\ln(x)}{(x^2 + 2x + 10)}$$

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. Jan 9, 2013

### Dick

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. Jan 10, 2013

### HallsofIvy

Staff Emeritus
My first thought was to let $u= x^2+ 2x+ 10$, the denominator, since its derivative is $2x+ 2= 2(x+ 1)$, but, of course, the "ln(x)" messes that up.

Okay so my second thought was to let $u= ln(x)$ so that $du= (1/x)dx$, $xdu= e^u du= dx$. Since $x= e^u$, the integral becomes
$$\int \frac{e^u+1}{e^{2u}+ 2e^u+ 1}e^udu= \int \frac{e^2u+ e^u}{e^{2u}+ 2e^u+ 1}du$$
and now try the substitution $v= e^{2u}+ 2e^{u}+ 1$.

10. Jan 10, 2013

### SteamKing

Staff Emeritus
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. Jan 10, 2013

### SammyS

Staff Emeritus
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. Jan 10, 2013

### Mandelbroth

Didn't the OP say we needed integration by parts?

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

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

I used the "\rightarrow" to indicate skipping the dreary use of partial fractions. I think you guys can take it from here...

13. Jan 10, 2013

### haruspex

It said and/or.
With the possible exception of the OP poster, I think we already did. . (I've never been sure whether 'OP' stands for Original Post or Original Poster, but I've always taken it to be the former.)