# Integration by partial Fractions question

## Homework Statement

The problem asks to evaluate the integral using partial fractions, but I just cannot find out which trick to get this one to work. the equation is
$$\int$$$$\frac{x^3+x^2+2x+1}{(x^2+1)(x^2+2)}$$

## The Attempt at a Solution

I have tried setting it up as a partial fraction with each part of the numerator under Ax+B and Cx+D but I have a feeling that is incorrect. Thanks in advance!

Dick
Homework Helper
Don't let a feeling it's incorrect stop you from doing it. If you followed through with your original program, you'd be done now. That's perfectly correct.

When I match up the coefficients to the problem i then get

A+C=1

B+D=1

2A+C=2

2B+D=1

This is where I get stuck again because how can B+D and 2B+D both equal 1?

Dick
Homework Helper
They can both be equal to 1 if B=0 and D=1. Don't overthink it. Just solve the equations. Subtract B+D=1 from 2B+D=1 and you'll reach that conclusion.

thanks...i always over think the problems and forget about 0....thanks so much. calc 2 is really kickin my butt

Dick
Homework Helper
You could kick it's butt instead. Everything you posted was exactly correct. Just push ahead and try it before you start thinking you must be wrong.

I'm having problems with another integral of similar properties.

The integral is
$$\int$$$$\frac{1}{x\sqrt{x+1}}$$

I made this into a u sub, where u = $$\sqrt{x+1}$$ and u^2 = x+1 Therefore x+u^2-1

So the integral is now $$\int$$$$\frac{1}{u^3-u}$$ which i made into $$\int$$$$\frac{1}{u(u-1)(u+1)}$$

I then used partial fractions to make it $$\frac{A}{u}$$ + $$\frac{B}{u+1}$$+$$\frac{C}{u-1}$$

Getting me A=-1 B=$$\frac{1}{2}$$ and C=$$\frac{1}{2}$$

Which changed to equation into $$\int$$$$\frac{-1}{u}$$+$$\frac{1}{2}$$$$\int$$$$\frac{du}{u+1}$$+$$\frac{1}{2}$$$$\int$$$$\frac{du}{u-1}$$

Which gets me to the answer -ln|$$\sqrt{x+1}$$|+ 1/2 ln|$$\sqrt{x+1}+1$$|+1/2 ln |$$\sqrt{x+1}-1$$|

but the answer is ln $$\frac{[tex]\sqrt{x+1}+1}{\sqrt{x+1}-1}$$

Can anyone provide me with help as to where I am messing up? Thanks

Mark44
Mentor
In your original integral you omitted dx. In your substitution, you neglected to calculate du, which is not equal to dx.

Thanks, I'll redo the equation right now
EDIT: ahhh I did it over with dx in there....thanks a bunch

I have to do

$$\int$$$$\frac{2x}{(x-3)^2}$$ with upper bound 2 and lower bound 0

What I've done so far:

Partial Fractions, making it: $$\frac{A}{x-3}$$+$$\frac{B}{(x-3)^2}$$

=> A(x-3)+B=2x
=> A=2 B=6
=> $$\int$$$$\frac{2}{x-3}$$ + $$\frac{6}{((x-3)^2)}$$
=> 2 ln|x-3| - 6 ln|t-3| again, evaluated from 0 to 2

I end up with -2 ln|3| + 6 ln|3|

The answer is 4-ln|9| Can anyone tell me where I'm going wrong? Thanks

Mark44
Mentor
$$\int \frac{6}{(x - 3)^2}dx \neq ln|(x - 3)^2| + C$$

Also, you could have done the problem more simply by using an ordinary substitution: u = x - 3, du = dx. You would have gotten u + some stuff in the numerator and u^2 in the denominator, which can be integrated easily.

Thanks with the hint on u substitution, that was a lot easier

Another question if possible.

How would you start the integral

$$\int$$$$\frac{earctany)}{1+y^2}$$

with boundries of -1 to 1

I can't seem to find the trick...

Last edited:
Mark44
Mentor
u = arctan(y).
What is du?

If you omit du, you're hosed.

Hint: pay attention to and don't lose those differentials!

ahhh yes i see it now. Thanks a bunch.

Can anyone help me as to where to start this one out? Thanks

$$\int$$(1+$$\sqrt{x}$$)^8 dx

Dick