# Def Integral of ln(x+2)dx

1. Oct 26, 2007

So I am using int by parts to evaluate $$\int_{-1}^1\ln(x+2)dx$$

I am doing something incorrectly with the indefinite int part of this problem, but I can'y find my error, can you?

$$\int udv=uv-\int vdu$$
$$u=ln(x+2)$$
$$\Rightarrow du=\frac{dx}{x+2}$$
$$dv=dx$$
$$\Rightarrow v=x$$

So I get:
$$x\ln(x+2)-\int\frac{x}{x+2}dx$$

Working on only the last part ^^^by long division I get:
$$\int\frac{x}{x+2}dx=\int(1-\frac{2}{x+2})dx$$

$$=x-2\ln(x+2)$$

and recombining with the top:

$$x\ln(x+2)-x+2\ln(x+2)$$

But when I evaluate this on my calculator, I get the incorrect answer. Any ideas?

Casey

Last edited: Oct 26, 2007
2. Oct 26, 2007

### mdk31

Why use integration by parts? Why not just use substiution?

u=x+2
du/dx=1

3. Oct 26, 2007

Because that is not the point of the exercise. A u-sub will do no good if we have not yet learned the integral of lnu

Casey

But, nonetheless, my integral is correct. I am doing something incorrectly with my calculator.

edit: I figured it out. damn. what a waste of typing

Last edited: Oct 26, 2007
4. Oct 26, 2007

5. Oct 26, 2007

### neutrino

The answer seems fine to me...

Use the distributive law to 'pull' the ln(x+2) out.

6. Oct 26, 2007

### sutupidmath

yeah i also got the same thing,

7. Oct 26, 2007

wow....yeah. The things I forget sometimes

Thanks guys.

BTW, it is common too use, as Wolfram does, LOG instead of LN....I thought that Log implied base 10 always?

8. Oct 26, 2007

### neutrino

To each his own. It's best to use what is being used in your course/textbook, at least for now.

9. Oct 26, 2007

### sutupidmath

Well, this depends pretty much on the author. Here at my country when we write only Log we assume it to be with base 10, and Ln to be the natural logarithm with base e. However i have seen some proffesors at some universities usually at US universities often use Log to mean Ln.

10. Oct 26, 2007

### ZioX

'Pure' mathematicians will generally have log denote the logarithm with base e. This is natural as all power functions (a^x) are really just scaled exponentials and we have a similar result for logs as a consequence.

11. Aug 1, 2008

### mj12

I 've been trying to work out an integral similar to the one above but i cant seem to get it correct
$$\int \ln(x+x^{2}) dx$$

$$u = \ln(x+x^{2})$$
$$du = dx/(x+x^{2})$$

$$dv = dx$$ : $$v = x$$

$$\int \ln(x+x^{2})= x \ln (x+x^{2})+ \int x /(x+x^{2})dx$$

To figure out the last part of the integral for some reason has me stuck :(

Last edited: Aug 1, 2008
12. Aug 1, 2008

### Defennder

13. Aug 1, 2008

### mj12

ok. but the differential of $$\ln(x) = 1/x$$

14. Aug 1, 2008

### Defennder

But the expression inside the ln is x^2+x and not x. So you have to apply the general rule for differentiating ln(f(x)).

15. Aug 1, 2008

### snipez90

mj12, you forgot the most important rule of differential calculus ;)

16. Aug 1, 2008

### mj12

By General rule you mean :ln f(x) = f`(x) / f(x)

which would give me 1 + 2x / x + x^2

17. Aug 1, 2008

### Defennder

Yes.

18. Aug 1, 2008

### snipez90

Ahhh, your notation had me confused for a second. If you mean (1+2x)/(x+x^2), you are correct. Now multiply that by x (to get vdu) and see if you can integrate that (it's a bit tricky).

19. Aug 1, 2008

### D H

Staff Emeritus
Try factoring $x+x^2[/tex] and then use the identity [itex]\log(ab) = \log a + \log b$.

20. Aug 1, 2008

### Defennder

That surprisingly, is the harder way to do it, mostly because it's easier to remember the anti-derivative of f'(x)/f(x) than that of ln(x).