# Integral of a product.

1. Oct 19, 2003

### PrudensOptimus

How do you find the integral of a product, say x*sqrt(1+x) dx...

I am very confuzed with all the notation is involved, can someone explain and do a few examples? Thanks.

2. Oct 19, 2003

### Tom Mattson

Staff Emeritus
"The Product Rule" for integration can be loosely said to be Integration by Parts. Actually, IBP is derived from the product rule for differentiation.

IBP is: &int;udv=uv-&int;vdu. One does this with the hope that &int;vdu is easier than &int;udv.

To do &int;x(1+x)1/2dx, let u=x and dv=(1+x)1/2dx.

Then du=dx and v=(2/3)(1+x)3/2, so and IBP says that:

&int;x(1+x)1/2dx=(2/3)x(1+x)3/2-(2/3)&int;(1+x)3/2dx

And the integral on the right hand side is easy to do.

Last edited: Oct 19, 2003
3. Oct 19, 2003

### HallsofIvy

Staff Emeritus
Yes, integration by parts is the "product rule in reverse".

However, for this example, I don't think that's what you would want to use. For the problem &int;x(x+1)1/2dx, it's not the product that is the problem: x(x1/2 would be easy enough: it's that "x+1" inside the root. I would make the easy substitution
u= x+1. Then du= dx. Of course, x is now u- 1 but that's easy- it's to the first power. x(x+1)1/2= (u+1)u1/2=
u3/2+ u1/2 so

&int;x(x+1)1/2dx= &int;(u3/2+ u1/2)du= (2/5)u5/2+ (2/3)u3/2+ C= (2/5)(x+1)5/2+ (2/3)(x+1)3/2+ C which, I think you will find, is exactly what Hurkyl's method gives.

4. Oct 19, 2003

### Sting

You could solve your integral by tabular integration.

[inte](x(1 + x)(1/2)dx

Essentially, you have two functions, x and (1 + x)(1/2)

Take the consecutive derivatives of x and consecutive integrals of
(1 + x)(1/2)

Column A
row 1: x
row 2: 1
row 3: 0

Column B
row 1: (1 + x)(1/2)
row 2: (2/3)(1 + x)(3/2)
row 3: (4/15)(1 + x)(5/2)

Now, I'll multiply column A, row 1, with column B, row 2. Then I'll subtract the product of column A, row 2 and column B and row 3 (subtraction simply because we alternate signs).

In tabular integration, you'll find the consecutive derivative of one function and the consecutive integrals of the other functions. It only works if you have a derivative that will eventually become zero and an integral that you can actually obtain.

So in our example, the integral will be:
x(2/3)(1 + x)(3/2) - (4/15)(1 + x)(5/2) + C

EDIT: syntax error

Last edited: Oct 19, 2003
5. Oct 20, 2003

Okay, so we have the tabular method, where [inte]vdu zero's out and we arrive at the the answer. Is there a name for the other case of IBP, where we will end up with some value 2([inte]udv)=(some known value ) ?

6. Oct 20, 2003

### PrudensOptimus

Can't it be something related to

d/dx (uv) = uv' + vu'??

since integral of uv is F(x), d/dx F(x) = uv

7. Oct 21, 2003

### KLscilevothma

I think what you are looking for is to use "Integration by parts" to do this question. Tom has already demonstrated it.

IBP means integration by parts.

8. Oct 21, 2003

### HallsofIvy

Staff Emeritus
"integration by parts" is basically the opposite of "the product rule" for differentiation.

The "tabular integration" mentioned above is a method of regularizing repeated integration by parts.