# Indefinite integral by partial fractions

• Mar 20th 2014, 11:14 AM
chalupabatman
Indefinite integral by partial fractions
Hey everyone I need some help with this problem that I have. I know that I put partial fractions into the title but I am unsure if that is truly what I need in order to complete this question. All my instructions says is:

Solve the following integral

This is it:

$\displaystyle \int\frac{{x}}{{(x-1)}{(\sqrt{x}-1)}}dx$

This is my method to solve:

$\displaystyle \int\frac{{x}}{{(x-1)}{(\sqrt{x}-1)}}dx = \frac{A}{x-1}+\frac{B}{\sqrt{x}-1}$

I multiplied both sides by the denominator of the left side:

$\displaystyle \int{x}dx = A(\sqrt{x}-1) + B(x-1)$

I factor in the A and B:

$\displaystyle x = A\sqrt{x} - A + Bx - B$

but this is where I am stuck at.. I mean the square root is where I am stuck at because the right and left sides aren't like terms right? Could definitely use some help with this one.

• Mar 20th 2014, 11:42 AM
romsek
Re: Indefinite integral by partial fractions
Quote:

Originally Posted by chalupabatman
Hey everyone I need some help with this problem that I have. I know that I put partial fractions into the title but I am unsure if that is truly what I need in order to complete this question. All my instructions says is:

Solve the following integral

This is it:

$\displaystyle \int\frac{{x}}{{(x-1)}{(\sqrt{x}-1)}}dx$

This is my method to solve:

$\displaystyle \int\frac{{x}}{{(x-1)}{(\sqrt{x}-1)}}dx = \frac{A}{x-1}+\frac{B}{\sqrt{x}-1}$

I multiplied both sides by the denominator of the left side:

$\displaystyle \int{x}dx = A(\sqrt{x}-1) + B(x-1)$

I factor in the A and B:

$\displaystyle x = A\sqrt{x} - A + Bx - B$

but this is where I am stuck at.. I mean the square root is where I am stuck at because the right and left sides aren't like terms right? Could definitely use some help with this one.

The key here is to recognize that

$(x-1) = (\sqrt{x}-1)(\sqrt{x}+1)$ and so your denominator is actually $(\sqrt{x}-1)^2(\sqrt{x}+1)$

Now rework it using the rules for that denominator.
• Mar 20th 2014, 11:44 AM
chalupabatman
Re: Indefinite integral by partial fractions
oh because you can technically factor the (x-1)... I didn't know that I will rework it and see what I can get
• Mar 20th 2014, 10:03 PM
hollywood
Re: Indefinite integral by partial fractions
I would substitute $\displaystyle u=\sqrt{x}$ and then use the normal partial fractions. I think you will be able to do the partial fraction expansion using romsek's hint, but then you might have trouble integrating it.

- Hollywood
• Mar 21st 2014, 05:36 AM
HallsofIvy
Re: Indefinite integral by partial fractions
Also, do NOT include the "$\displaystyle \int$" in your calculations for the partial fractions. Writing "u" for $\displaystyle \sqrt{x}$ as Hollywood suggests, that would be
$\displaystyle \frac{u^2}{(u- 1)^2(u+ 1)}= \frac{A}{u+1}+ \frac{B}{u- 1}+ \frac{C}{(u- 1)^2}$
• Mar 21st 2014, 06:51 AM
hollywood
Re: Indefinite integral by partial fractions
I was thinking of actually making a u-substitution in the integral, so setting $\displaystyle u=\sqrt{x}$ and $\displaystyle du=\frac{1}{2\sqrt{x}}\,dx$:

$\displaystyle \int\frac{x}{(x-1)(\sqrt{x}-1)}\,dx =$

$\displaystyle \int\frac{x}{(x-1)(\sqrt{x}-1)}2\sqrt{x}\frac{1}{2\sqrt{x}}\,dx =$

$\displaystyle \int\frac{u^2}{(u^2-1)(u-1)}2u\,du =$

$\displaystyle \int\frac{2u^3}{(u+1)(u-1)^2}\,du$

Then expand:

$\displaystyle \frac{2u^3}{(u+ 1)(u- 1)^2}= \frac{A}{u+1}+ \frac{B}{u-1}+ \frac{C}{(u-1)^2}$

- Hollywood
• Mar 21st 2014, 09:06 AM
chalupabatman
Re: Indefinite integral by partial fractions
Quote:

Originally Posted by HallsofIvy
Also, do NOT include the "$\displaystyle \int$" in your calculations for the partial fractions. Writing "u" for $\displaystyle \sqrt{x}$ as Hollywood suggests, that would be
$\displaystyle \frac{u^2}{(u- 1)^2(u+ 1)}= \frac{A}{u+1}+ \frac{B}{u- 1}+ \frac{C}{(u- 1)^2}$

So let me see if I got this one right.. you factored the $\displaystyle (x-1)$ then you used the substitution for u = x?
• Mar 21st 2014, 09:13 AM
chalupabatman
Re: Indefinite integral by partial fractions
I tried again and used u substitution this is my new process but I still think it is wrong:

$\displaystyle u^{2} = x$
$\displaystyle 2udu = dx$

Then I subbed into my equation to get:

$\displaystyle \frac{u^{2}*2udu}{(u^{2}-1)(u-1)}$

Then I proceeded with the partial fractions:

$\displaystyle \frac{u^{2}*2udu}{(u^{2}-1)(u-1)} = \frac{A}{(u^{2}-1)} + \frac{B}{(u-1)}$

Multiplied both sides by the denom:

$\displaystyle u^{2}*2udu = A (u^{2}-1) + B (u-1)$

but when I solve this I get:

$\displaystyle B = 1, A = 2$

which really doesn't really make sense to me.. I will try it your way to see how I fair but I just wanted to update you guys with my work and progress
• Mar 21st 2014, 09:49 AM
romsek
Re: Indefinite integral by partial fractions
Look at post #6 again. Hollywood has almost done it all for you. You didn't use the correct denominator. It's $(u-1)^2$ not $(u^2-1)$.
• Mar 27th 2014, 12:10 PM
chalupabatman
Re: Indefinite integral by partial fractions
Hey everyone sorry for the long reply I had a death in my family but I am still trying to get this one done... I did what post #6 did, but when I showed my professor he said that it was wrong because it is the same degree on the top and bottom when you factor out the denominator

$\displaystyle \frac{2u^{3}}{u^{3}-u^{2}-u+1}$

So he informed me to do long division which I did and I got this:

$\displaystyle \frac{2 * \frac{u^{2}+2u-2}{u^{3}-u^{2}-u+1}}{u^{3}-u^{2}-u+1}$

then this turned into:

$\displaystyle \frac{2(u^{2}+2u-2)}{(u^{3}-u^{2}-u+1)^{2}}$

which I factored to this:

$\displaystyle \frac{2(u^{2}+2u-2)}{(u-1)^{4}(u+1)^{2}}$

I think I did something wrong because then wouldn't I have this long partial fractions with 6 different coefficients? Like:

$\displaystyle \frac{2(u^{2}+2u-2)}{(u-1)^{4}(u+1)^{2}} = \frac{A}{(x+1)} + \frac{B}{(x+1)^{2}} + \frac{C}{(x-1)} \frac{D}{(x-1)^{2}} \frac{E}{(x-1)^{3}} \frac{F}{(x-1)^{4}}$

It just doesn't really make sense to me.. thanks in advance guys
• Mar 28th 2014, 07:05 AM
hollywood
Re: Indefinite integral by partial fractions
Your professor is correct, you need to do long division first. The denominator needs to have a higher degree than the numerator for the partial fractions expansion to work. So you need to do:

$\displaystyle \frac{2u^3}{u^3-u^2-u+1} = 2 + \frac{u^2+u-1}{u^3-u^2-u+1} = 2 + \frac{u^2+u-1}{(u+1)(u-1)^2}$

Now the denominator has higher degree than the numerator so you can do the partial fraction expansion:

$\displaystyle \frac{u^2+u-1}{(u+1)(u-1)^2} = \frac{A}{u+1}+ \frac{B}{u-1}+ \frac{C}{(u-1)^2}$

- Hollywood
• Mar 30th 2014, 01:58 PM
chalupabatman
Re: Indefinite integral by partial fractions
Quote:

Originally Posted by hollywood
Your professor is correct, you need to do long division first. The denominator needs to have a higher degree than the numerator for the partial fractions expansion to work. So you need to do:

$\displaystyle \frac{2u^3}{u^3-u^2-u+1} = 2 + \frac{u^2+u-1}{u^3-u^2-u+1} = 2 + \frac{u^2+u-1}{(u+1)(u-1)^2}$

Now the denominator has higher degree than the numerator so you can do the partial fraction expansion:

$\displaystyle \frac{u^2+u-1}{(u+1)(u-1)^2} = \frac{A}{u+1}+ \frac{B}{u-1}+ \frac{C}{(u-1)^2}$

- Hollywood

BRILLIANT!! I have no idea what I was thinking hahaha.. thank you my dood I will be sure to work through this! Thanks again dood