Integrating by partial fractions

[suspenc3]

Hi, I'm having quite a bit of trouble with this topic. Here's one of the first problems, I don't really understand the method in the book, if someone could show me an easy route, it would help.

$$\int_{0}^{1} \frac {2x+3}{(x+1)^2}dx$$

Thanks

 

[0rthodontist]

It's faster not to use partial fractions in this case. You can let u = x + 1.

What don't you understand about the method in the book?

 

[arildno]

Notice that your integrand may be rewritten as:
$$\frac{2x+3}{(x+1)^{2}}=\frac{2x+2}{(x+1)^{2}}+\frac{1}{(x+1)^{2}}=\frac{2}{(x+1)}+\frac{1}{(x+1)^{2}}$$

 

[suspenc3]

I just find that the examples don't really tell me that much, I understand what they do in the examples but then when I get to a question, I rarely never know how to start it.

arildno, is there a certain way that you came up with that, or do you just see it?

 

[arildno]

The certain way is 3=2+1
And yes, I "just see it". (As you should as well)

 

[suspenc3]

ok, so for this one:

$$\int \frac{x^2}{(x-3)(x+2)^2}dx$$

would I start out like this:

$$\frac{A}{x-3} + \frac{B}{x+2} + \frac{C}{(x+2)^2}$$

and then they would just be 1, x, x

 

[arildno]

Eeh?
What do you mean by 1,x,x??

 

[0rthodontist]

A, B, and C must be constants (they can't be x). And you know one further thing: The sum you have there must be equal to the original fraction, i.e.
$$\frac{A}{x-3} + \frac{B}{x+2} + \frac{C}{(x+2)^2} = \frac{x^2}{(x-3)(x+2)^2}$$

Multiply it out, and then you know something else: for this to be true for ALL x the constant terms must be equal on both sides, and so must the coefficients of x, and so must the coefficients of x^2. You will then have three linear equations to solve--namely those that equate the coefficients of each power of x.

An alternative way that can help you get A, B, and C more rapidly is that you can pick x to be anything you want in that equation, since the equation must be true for all x. So what happens if you multiply both sides by (x - 3) and then set x = 3?

 

[suspenc3]

ok, i think I am getting that part..I have another problem now, I am not sure if I am going about it the right way.

$$\int_{0}^{1} \frac{x}{x^2 +4x +13}dx$$

=$$\int_{0}^{1}1+[\frac{-x^2-3x-13}{x^2 +4x +13}]dx$$..
in the book there is an example somewhat like this..they complete the square of the bottom..thats what i did and then substituted..i ended up with this so far:

$$x + \int \frac{-u^2+u-11}{u^2+9}du$$

if this is right so far, do i just split this up into several fractions and then integrate, or is there more?

 

[0rthodontist]

I am not familiar with the technique your book uses. Why did you split off the x? With u = x + 2, you could just turn your original integral into
$$\int_{2}^{3} \frac{u-2}{u^2 + 9}dx$$
which isn't too hard to integrate.

 

[arildno]

Ortho: Not dx, but du..

 

[suspenc3]

so you can do that?its that easy?..why from 3 to 2?

 

[suspenc3]

this isn't the example..this is a question that resembles the example..this is the example..$$\int \frac{4x^2-3x+2}{4x^2-4x+3}$$

 

[0rthodontist]

Ortho: Not dx, but du..

Oh yeah, missed that.

suspenc3, it's from 2 to 3 because when you change from x to u(x), you change the limits from (a to b) to (u(a) to u(b)).

 

[HallsofIvy]

The original problem was
$$\int_{0}^{1} \frac{x}{x^2 +4x +13}dx$$
Completing the square, x²+ 4x+ 13= (x+ 2)²+ 9. If you let u= x+ 2, then x= u- 2 so the integral becomes
$$\int_2^3\frac{u-2}{u^2+1}du$$

To integrate that, separate it as
$$\int_2^3 \frac{u du}{u^2+1}+ \int_2^3\frac{du}{u^2+1}$$
The first is an easy v= u²+ 1 substitution and the second is an arctangent.

I don't see any reason to take out that "1" in
$$\int_{0}^{1}1+[\frac{-x^2-3x-13}{x^2 +4x +13}]dx$$

I wonder if the example you looked at didn't have a higher power of x in the numerator equal to that in the denominator, say
$$\int_{0}^{1} \frac{x^2}{x^2 +4x +13}dx$$.
Since to use partial fractions you have to have the denominator with degree larger than the numerator, the first thing you would do with that is divide to get 1+ ...