# Integration by parts

1. Oct 19, 2005

### zenity

Hi,

I'm a bit confused as to what I should assign u and dv in this integration by parts:

ln(1+x^2)dx

I remember a general rule called the "LIPATE" rule... which is basically Logarithms, inverse trigs, poly, algebra, trig, then exponentials...

Now... would I assign u = ln(1+x^2)? and dv = dx?

That kind of defeats the purpose of integration by parts... doesn't it?

I'm probably not assigning u and dv properly... so if anybody can lend me a hand, thank you!

2. Oct 19, 2005

### Tom Mattson

Staff Emeritus
Yes.

No, it works like a charm. Try it and see.

3. Oct 20, 2005

### zenity

Ok... so this is what happens after I work it out

u=ln(1+x)^2 du=2x/(x^2+1)

dv=dx v=x

ln(1+x^2)x - integral [ 2x^2/(1+x^2)

Using a calc, I found out that the integral works out to be x - arctan(x)

... but I don't know how to do it ... any pointers?

4. Oct 20, 2005

### Tom Mattson

Staff Emeritus
Your "new" integrand is an improper rational function. They should have beat in into your head during your precalc courses that improper rational functions should be decomposed into a polynomial plus a proper rational function, via long division.

Try that, and then see if it's not clear what to do.

5. Oct 20, 2005

### zenity

I'll need a quick refresher on that... I don't remember exactly how to do that...
I'll take a look around on the net. In the meantime, I am 100% stuck on this question:

integral of x^7/(1+x^4)^(3/2)

I've tried assigning u=x^7 and dv="the bottom part"

unfortunately... I don't know where I'm headed.

I'm sorry for all these questions. I have a hard time trying to determine what I should assign to u and dv. Is there a good tutorial online or something that can help me in this area?

6. Oct 21, 2005

### Tom Mattson

Staff Emeritus
Don't get so caught up in integration by parts that you forget the other stuff you used to know. This can be done with a simple u-substitution. I'll let you figure out what u is supposed to be.

7. Oct 22, 2005

### zenity

Thanks... but this section in the book insists that we use integration by parts in order to solve these questions... (Although using substitution as a secondary method would be acceptable I think).

I've solved the questions... but I'm now stuck on integration and trigonometric identities...

Integral of sec(x)^4 / tan(x)^2

I've tried using sec(x)^2 = tan(x)^2 + 1 and vice versa, but I don't seem to be getting anywhere with that. If someone could possibly lend a few hinters! Thanks!

8. Oct 23, 2005

### VietDao29

You can just change everything to sin(x) and cos(x), then go from there.
So:
$$\int \frac{\sec ^ 4 x}{\tan ^ 2 x} dx = \int \frac{\cos ^ 2 x}{\sin ^ 2 x \cos ^ 4 x}dx = \int \frac{1}{\sin ^ 2 x \cos ^ 2 x}dx$$
There's sin(x) and cos(x) in the denominator, so you may want to change the numerator a bit so that it also has sin(x) and cos(x).
$$... = \int \frac{\cos ^ 2 x + \sin ^ 2 x}{\sin ^ 2 x \cos ^ 2 x}dx = ...$$
Can you go from here?
--------------------
Or if you want to continue doing it your way, then you can change everything into tan(x).
So:
$$\int \frac{\sec ^ 4 x}{\tan ^ 2 x} dx = \int \frac{(1 + \tan ^ 2 x) ^ 2}{\tan ^ 2 x} dx = \int \frac{dx}{\tan ^ 2 x} + \int \tan ^ 2 x dx + \int 2 dx$$
Now you can change tan(x) to sin(x) and cos(x): tan(x) = sin(x) / cos(x). And use sin2x + cos2x = 1 to solve the problem.
---------------
Anyway, it's always good that you start another thread for another problem, instead of using the old one...
Viet Dao,

Last edited: Oct 23, 2005