How can I verify the solution to this integration problem using division?

[lionely]

Homework Statement

Verify, by division, that 2x/(3x+1) = 2/3 - 2/3(3x+1)

Hence, evaluate ∫2x/(3x+1) dx


I don't understand what to, does the question mean to do long division?

Help is much appreciated!

 

[CompuChip]

It's easier to start from the right hand side: take 2/3 - 2/3(3x + 1) and combine the fractions into one. After simplification, you should get 2x/(3x + 1).

 

[vanhees71]

\frac{2x}{3x+1}=\frac{2}{3} \cdot \frac{3x}{3x+1}=\frac{2}{3} \cdot \frac{3x+1-1}{3x+1}=\ldots

 

[lionely]

Oh thanks, vanhees. But is that still division?

 

[Ray Vickson]

Oh thanks, vanhees. But is that still division?

"Division" means re-writing the expression so that in the remainder the numerator has a smaller degree than the denominator. So, if we have f(x) = p(x)/q(x), it is already in "divided" form if deg(p) < deg(q); otherwise (if deg(p) ≥ deg(q)), manipulate the expression to get f(x) = m(x) + r(x)/q(x), with deg(r) < deg(q). (Of course, I mean that p,q,m,r are all polynomials in x.) In your example, deg(p) = 1 = deg(q), so you need to re-write the expression until the remainder has numerator of degree 0 (that is, has the form c/(3x+1) for constant c).

 

[lionely]

Thank you Ray.

 

[arildno]

Or, by polynomial long division

2x:(3x+1)->2/3
2x+2/3
0-2/3

Implies:
2x/(3x+1)=2/3-2/3*(1/(3x+1))

 

[vanhees71]

\frac{2x}{3x+1}=\frac{2}{3} \cdot \frac{3x}{3x+1}=\frac{2}{3} \cdot \frac{3x+1-1}{3x+1}=\frac{2}{3} \left (1-\frac{1}{3x+1} \right).
Written in this form, the integral is very easy. I don't know, how you call these simple manipulations.

 

[Ray Vickson]

\frac{2x}{3x+1}=\frac{2}{3} \cdot \frac{3x}{3x+1}=\frac{2}{3} \cdot \frac{3x+1-1}{3x+1}=\frac{2}{3} \left (1-\frac{1}{3x+1} \right).
Written in this form, the integral is very easy. I don't know, how you call these simple manipulations.

It is impossible to tell which message you are responding to, since you did not use the "Quote" button. Anyway, these manipulations are called simple because they are simple. (However, as far as I can tell, nobody called them simple before you did, so I don't understand your statement!)

 

[vanhees71]

Come on, it is really simple to integrate now!

 

[CompuChip]

Euh, guys... am I the only one who is getting confused here, or are you all replying to the wrong messages. The OT seems to have gotten the hint back in post #6; none of the other posters is the topic starter and knows perfectly well how to solve the exercise, so what is the discussion about?

 

[arildno]

I merely offered an alternative solution procedure!

 

[D H]

Euh, guys... am I the only one who is getting confused here, or are you all replying to the wrong messages.

That's typical in homework problems. We have a tendency to drag the discussion off topic. The OP hasn't come back since post #6, so presumably the problem has been solved.

That said, I suspect arildno's post #7 is exactly what the instructor intended with the phrase "verify, by division."