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).