MHB What is the best method for solving rational function integrals?

[Yankel]

Hello,

I am trying to solve the next integral, using this method of rational functions.

My problem is, that I turn the function into two expressions with the same denominator...when I try to use values of x, I will always make zero of both sides, if you know what I mean

\[\int \frac{x+1}{(x-4)^{2}}dx=\int (\frac{A}{x-4}+\frac{B}{x-4})dx\]

I'll be happy to see an example how to solve such integrals.

Thank you !

 

[chisigma]

Hello,

I am trying to solve the next integral, using this method of rational functions.

My problem is, that I turn the function into two expressions with the same denominator...when I try to use values of x, I will always make zero of both sides, if you know what I mean

\[\int \frac{x+1}{(x-4)^{2}}dx=\int (\frac{A}{x-4}+\frac{B}{x-4})dx\]

I'll be happy to see an example how to solve such integrals.

Thank you !

The appropriate partial fraction expansion in this case is...

$\displaystyle \frac{x+1}{(x-4)^{2}} = \frac{A}{x-4} + \frac{B}{(x-4)^{2}}\ (1)$

Kind regards

$\chi$ $\sigma$

 

[I like Serena]

Hello,

I am trying to solve the next integral, using this method of rational functions.

My problem is, that I turn the function into two expressions with the same denominator...when I try to use values of x, I will always make zero of both sides, if you know what I mean

\[\int \frac{x+1}{(x-4)^{2}}dx=\int (\frac{A}{x-4}+\frac{B}{x-4})dx\]

I'll be happy to see an example how to solve such integrals.

Thank you !

How about:
$$\frac{x+1}{(x-4)^{2}} = \frac{(x-4)+5}{(x-4)^{2}}
= \frac{1}{x-4}+\frac{5}{(x-4)^{2}}$$

Logical follow-up is a substitution of $u=x-4$ (also good to start with).

 

[Yankel]

chisigma,

why ? I don't get it, if you multiply both fractions, you get (x-4)^3

 

[I like Serena]

why ? I don't get it, if you multiply both fractions, you get (x-4)^3

Try it like this:
$$\frac{A}{x-4} + \frac{B}{(x-4)^{2}}
= \frac{A(x-4)}{(x-4)^{2}} + \frac{B}{(x-4)^{2}}
= \frac{A(x-4) + B}{(x-4)^{2}}$$

 

[Fantini]

I hope you guys don't mind me invading the thread for a second.

chisigma,

why ? I don't get it, if you multiply both fractions, you get (x-4)^3

I think you are misunderstanding partial fractions. :) If you had

$$\frac{1}{3} + \frac{1}{9}$$

the result would be

$$\frac{1}{3} + \frac{1}{9} = \frac{3}{9} + \frac{1}{9} = \frac{4}{9}.$$

The same happens with polynomials. By writing

$$\frac{A}{x-4} + \frac{B}{(x-4)^2}$$

what you mean is

$$\frac{A}{x-4} + \frac{B}{(x-4)^2} = \frac{A(x-4)}{(x-4)^2} + \frac{B}{(x-4)^2} = \frac{A(x-4) + B}{(x-4)^2}.$$

Guess all I am trying to say is that the common denominator is $(x-4)^2$ and not $(x-4)^3$. In other words, you don't multiply them by each other, because they have a factor in common (one $(x-4)$). ;)

Cheers!

 

[Yankel]

Right, now that I have seen the analogy to number I get it...looks obvious now

 

[DreamWeaver]

Hello,

I am trying to solve the next integral, using this method of rational functions.

My problem is, that I turn the function into two expressions with the same denominator...when I try to use values of x, I will always make zero of both sides, if you know what I mean

\[\int \frac{x+1}{(x-4)^{2}}dx=\int (\frac{A}{x-4}+\frac{B}{x-4})dx\]

I'll be happy to see an example how to solve such integrals.

Thank you !

This is so, so, so much easier if you just represent the x in the numerator as a function of the denominator - namely, $$x=(x-4) +4\,$$ . Observe:

$$\int \frac{x+1}{(x-4)^{2}}dx=\int\frac{[(x-4)+4]+1}{(x-4)^2}\,dx=$$

$$\int \frac{(x-4)+5}{(x-4)^2}\,dx=\int\frac{dx}{(x-4)}+5\int\frac{dx}{(x-4)^2}\,dx$$Hope that helps... :D