How Do You Integrate (u^4 + 1) / (u^5 + 3u)?

[scottshannon]

Homework Statement

Integral (u^4 +1) du/(u^5+3u)

Homework Equations

I don't know how to solve this integral. I don't know how to simplify it or what to do.

The Attempt at a Solution

It is not a form of dt/t. I tried to substitute trig functions and got nowhere. I thought it might be able to be solved by partial fractions but no.

 

[LCKurtz]

Homework Statement

Integral (u^4 +1) du/(u^5+3u)

Homework Equations

I don't know how to solve this integral. I don't know how to simplify it or what to do.

The Attempt at a Solution

It is not a form of dt/t. I tried to substitute trig functions and got nowhere. I thought it might be able to be solved by partial fractions but no.

Partial fractions is in fact a good idea, although a little tricky. Here's a hint. Try$$
\frac{u^4+1}{u^5+3u}=\frac{u^4+1}{u(u^4+3)}=\frac A u +\frac {Bu^3}{u^4+3}$$

 

[scottshannon]

Thank you. I will see how that works, although I probably would have tried Bu^4+ Cu^3 +Du^2 +Eu +F. How do you know C, D, E and F = 0? I am asking myself this question.

 

[PeroK]

Thank you. I will see how that works, although I probably would have tried Bu^4+ Cu^3 +Du^2 +Eu +F. How do you know C, D, E and F = 0? I am asking myself this question.

Perhaps LCKurtz is cleverer than either of us, but I would have made it easier than you by doing:

$\frac{A}{u} + \frac{X}{u^4 + 3}$ where $A$ is a constant and $X$ is a polynomial in $u$. You can then solve for $X$ without all the variables for each coefficient. Perhaps then half way through this calculation, like me, you see the shortcut that LCK saw.

 

[Ray Vickson]

Thank you. I will see how that works, although I probably would have tried Bu^4+ Cu^3 +Du^2 +Eu +F. How do you know C, D, E and F = 0? I am asking myself this question.

In general, a partial fraction for
$$f(u) = \frac{u^4+1}{u(u^4+3)}$$
would have the form of a sum of terms like $P(u)/Q(u)$, where the degree of $P$ is at least one less than the degree of $Q$. Thus
$$\begin{array}{rcl} \displaystyle \frac{u^4+1}{u(u^4+3)} &=& \displaystyle \frac{A}{u} + \frac{Bu^3 + Cu^2 + Du+E}{u^4+3}\\ & =& \displaystyle \frac{A(u^4+3) + u(Bu^3 + Cu^2 + Du+E)}{u(u^4+3)} \end{array}$$
The numerator above is a polynomial of degree 4 that must be equal to $u^4+1$ for all u, so you can determine unique values of $A,B,C,D,E$ from that.

 

[LCKurtz]

Partial fractions is in fact a good idea, although a little tricky. Here's a hint. Try$$
\frac{u^4+1}{u^5+3u}=\frac{u^4+1}{u(u^4+3)}=\frac A u +\frac {Bu^3}{u^4+3}$$

Thank you. I will see how that works, although I probably would have tried Bu^4+ Cu^3 +Du^2 +Eu +F. How do you know C, D, E and F = 0? I am asking myself this question.

I wasn't sure it would work, but noticing that the numerator had only $u^4$ and a constant term, and that my proposed partial fractions would generate only terms like that, it seemed worth trying it. Had it not worked, one of the methods others have suggested would have been necessary.

 

[SammyS]

Partial fractions is in fact a good idea, although a little tricky. Here's a hint. Try$$
\frac{u^4+1}{u^5+3u}=\frac{u^4+1}{u(u^4+3)}=\frac A u +\frac {Bu^3}{u^4+3}$$

A way to get this: (not necessarily any better than what's been said)

$\displaystyle \frac{u^4+1}{u^5+3u}$

$\displaystyle = \frac{3u^4+3}{3u(u^4+3)}$

$\displaystyle = \frac{2u^4 +u^4+3}{3u(u^4+3)}$

$\displaystyle = \frac{2u^4}{3u(u^4+3)}+ \frac{u^4+3}{3u(u^4+3)}$

$\displaystyle = \frac{2u^3}{3(u^4+3)}+ \frac{1}{3u}$​