# Integration by partial Fractions question

• bns1201
In summary, the problem asks to evaluate the integral using partial fractions, but the equation is giving me trouble. TheAttempt at a Solution gave a partial fraction decomposition of the integral and the problem was solved by plugging in the coefficients.
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!

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?

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

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

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

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

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:
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

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

Last edited:
thanks again I appreciate it

## What is integration by partial fractions?

Integration by partial fractions is a technique in calculus used to simplify and solve integrals involving rational functions. It involves breaking down a complex rational function into simpler fractions that can be easily integrated.

## When is integration by partial fractions used?

Integration by partial fractions is used when integrating rational functions, which are functions in the form of a polynomial divided by another polynomial. It is especially useful when the degree of the numerator is less than the degree of the denominator.

## How do you perform integration by partial fractions?

To perform integration by partial fractions, you first need to factor the denominator of the rational function into linear and irreducible quadratic factors. Then, using the method of undetermined coefficients, you can determine the coefficients of the simpler fractions. Finally, you can integrate each fraction separately to obtain the final result.

## What are the benefits of using integration by partial fractions?

One of the main benefits of using integration by partial fractions is that it allows for easier integration of rational functions. It also helps to simplify complex integrals into simpler ones, making it easier to evaluate them. Additionally, it can be used to solve problems in various fields such as physics, engineering, and economics.

## Are there any limitations to integration by partial fractions?

Integration by partial fractions is limited to rational functions only. It cannot be used to integrate other types of functions such as trigonometric or exponential functions. It is also not applicable to improper integrals, where the limits of integration are infinite.

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