Integration by partial Fractions question

[bns1201]

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]

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.

 

[bns1201]

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]

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.

 

[bns1201]

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

 

[Dick]

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.

 

[bns1201]

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{\sqrt{x+1}+1}{\sqrt{x+1}-1}<br /> <br /> <br /> <br /> <br /> <br /> Can anyone provide me with help as to where I am messing up? Thanks

 

[Mark44]

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

 

[bns1201]

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

 

[bns1201]

Another question please:
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]

Your partial fraction decomposition was fine, but your 2nd integral wasn't.
\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.

 

[bns1201]

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


Another question if possible.

How would you start the integral

\int\frac{e^arctany)}{1+y^2}

with boundries of -1 to 1

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

 

[Mark44]

u = arctan(y).
What is du?

If you omit du, you're hosed.

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

 

[bns1201]

ahhh yes i see it now. Thanks a bunch.

 

[bns1201]

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


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

 

[Dick]

Start with the obvious and unavoidable u=1+sqrt(x). Where does that take you?

 

[bns1201]

thanks again I appreciate it