Tricky rational function integral

[terhorst]

Homework Statement

$$\int \frac{4x^5-1}{(x^5+x+1)^2} dx = ?$$​

Homework Equations

The solution is

$$- \frac{x}{x^5 + x + 1}$$​

The Attempt at a Solution

Other than getting lucky and noticing immediately that this could be the derivative of a fraction, I do not see an easy way to solve this. The partial fractions solution is long and involved. I am reviewing calculus material to study for the math GRE, so am trying to see if there is a more efficient way. Given the level of the other exercises in this book, I would be surprised if there were not. Help me see the light! Thanks :-)

 

[Tedjn]

I don't know a more systematic way than partial fractions. Of course, you might look at (x⁵ + x + 1)' = 5x⁴ + 1 and observe that x(5x⁴ + 1) - (x⁵ + x + 1) gives you the cancellation you need. You might even suspect it from looking at how convoluted the solution would be otherwise. Still, not so easy without hindsight. If there is indeed a better way to handle such questions, I too would like to know.

 

[vela]

This is a problem I would have skipped if I had seen it on the math GRE. :) Since the GRE is multiple choice, you could always take the answers and differentiate them and see if you recover the integrand.

Generally, though, I'm not sure how you'd solve this problem. The squared function on the bottom suggests that it could be the result of the quotient rule. That might be enough of a hint to notice what Tedjn said.

 

[Dick]

You can do it by parts. Use v=1/(x^5+x+1) with the original integral being u*dv. It just happens that the v*du integral works out nicely. It's not a general method. It just happens to work out that way. And I only guessed that by working backwards from the answer. I really can't see any reason to put this on the GRE, except for cruelty.

 

[njama]

Here is an efficient way:
$$\frac{4x^5-1}{(x^5+x+1)^2} = \frac{Ax^4+B}{x^5+x+1}+\frac{Cx^4+D}{(x^5+x+1)^2}$$

$$4x^5-1 = (Ax^4+B)(x^5+x+1)+(Cx^4+D)$$

I think it's much faster then the other methods.

 

[HallsofIvy]

Multiplying that out,
$$4x^5- 1= Ax^9+ (A+ B)x^5+ Cx^4+ Bx+ A+ B+ D$$
Since there is no $x^9$ term on the left, A= 0. Then A+ B= B= 4, but there is no "4x" term on the left. That does not work. There are no "A", "B", "C", and "D" satisfying that equation for all x.