# Integral of a product.

#### 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.

#### Tom Mattson

Staff Emeritus
Gold Member
"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.

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#### HallsofIvy

Homework Helper
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.

#### 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

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#### On Radioactive Waves

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 ) ?

#### 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

#### KLscilevothma

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

originally posted by Tom
IBP is: &int;udv=uv-&int;vdu. One does this with the hope that &int;vdu is easier than &int;udv.
IBP means integration by parts.

#### HallsofIvy

Homework Helper
"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.

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