# Problem with these types of integrals

1. Mar 19, 2006

### Jacob87411

For some reason I just have a problem with these types of integrals..if someone could show me how to do one it would save me a lot of headaches..

1) x+2/(x^2-4) from 0 to 1

This just simplifies to 1/(x-2) so this should be ln(1-2)-ln(0-2)=ln(-1)-ln(-2)?

2) x/x+1, x/x^2+2...Do you use by parts? I cannot get any of these right where its of this nature..thanks so much

2. Mar 19, 2006

### EbolaPox

For number one, you integrated correctly. However, remember your logarithm rules.

$$ln(a) - ln(b) = ln(a/b)$$

Here's a hint for two: If you have something such as$$\int \frac{x dx}{x+a} = \int \frac{x+a-a dx}{x+a} = \int 1dx - \int \frac{a dx}{x+a}$$

You can generalize that further if you have the case x+ k on top. I think that may help.

Furthermore, on the x/x^2 + 2, try u-substitution.

Last edited: Mar 19, 2006
3. Mar 20, 2006

### VietDao29

This is wrong!!!
You are forgetting the absolute value. You should note that in the reals, ln(x) is only defined for positive x. Hence, there's no such thing as ln(-1), or ln(-2).
You can look again at the table of integrals, you may find something that reads:
$$\frac{dx}{x} = \ln |x| + C$$
For the first one, as EbolaPox has pointed out, you should split it into 2 integrals:
$$\int \frac{x}{x + 1} dx = \frac{x + 1 - 1}{x + 1} dx = \int dx - \int \frac{dx}{x + 1}$$.
For the second one, if your numerator is 1 degree less than your denominator, then it's common to try to take the derivative of the denominator, then rearrange the numerator so that it contains a multiple of the derivative of the denominator. Your answer will have a ln part and an arctan part (sometimes it does not have an arctan part).
$$\int \frac{x}{x ^ 2 + 2} dx$$
Now:
(x2 + 2)' = 2x. Now try to do some manipulation on the numerator to get a multiple of 2x:
$$\int \frac{x}{x ^ 2 + 2} dx = \int \frac{\frac{1}{2} \times 2x}{x ^ 2 + 2}$$
Now use the u-substitution:
u = x2 + 2, we have:
$$\int \frac{x}{x ^ 2 + 2} dx = \frac{1}{2} \int \frac{du}{u} = \frac{1}{2} \ln |u| + C = \frac{1}{2} \ln |x ^ 2 + 2| + C = \frac{1}{2} \ln (x ^ 2 + 2) + C$$.
Since x2 + 2 is positive for all x, so:
|x2 + 2| = x2 + 2.
Can you get it? :)