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:

This is my method to solve:

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

I factor in the A and 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.

Thanks in advance

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:

This is my method to solve:

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

I factor in the A and 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.

Thanks in advance

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.

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

Re: Indefinite integral by partial fractions

I would substitute 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

Re: Indefinite integral by partial fractions

Also, do NOT include the " " in your calculations for the partial fractions. Writing "u" for as Hollywood suggests, that would be

Re: Indefinite integral by partial fractions

I was thinking of actually making a u-substitution in the integral, so setting and :

Then expand:

- Hollywood

Re: Indefinite integral by partial fractions

Quote:

Originally Posted by

**HallsofIvy** Also, do NOT include the "

" in your calculations for the partial fractions. Writing "u" for

as Hollywood suggests, that would be

So let me see if I got this one right.. you factored the then you used the substitution for u = x?

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:

Then I subbed into my equation to get:

Then I proceeded with the partial fractions:

Multiplied both sides by the denom:

but when I solve this I get:

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

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 $(u^2-1)$.

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

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

then this turned into:

which I factored to this:

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

It just doesn't really make sense to me.. thanks in advance guys

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:

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

- Hollywood

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:

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

- 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