Find term of a rational partial (reduced) fraction

[sotownsend]

Homework Statement

Find the coefficient on the x^2 term in ∫[(x+9)/3] * [log(x+1)] * dx

Answer: -1/12

Homework Equations

Integration by parts

∫dv * u = u * v - ∫du * v

The Attempt at a Solution

1. I integrated by parts to get
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∫[(x+9)/3] * [log(x+1)] * dx] = (x^2/9)*log(x+1) - ∫[(s^2 + 18s)/(6s+6)]

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I think the term the problem wants something with the ∫[(s^2... term, but I don't know how to get the coefficient from it.

The problem suggests that ∫[(s^2 + 18s)/(6s+6)] is "now a rational function problem: divide to get a polynomial part and a reduced fraction. The reduced fraction gives a log term."

I very much appreciate any help!

 

[Stephen Tashi]

I think the term the problem wants something with the ∫[(s^2... term, but I don't know how to get the coefficient from it.

Forget how to divide a polynomial into a polynomial? I'll have to say it was a topic in elementary algebra that I did paid much attention to either. $6s + 6$ goes into $s^2 + 18s$ with quotient $\frac{s}{6} + \frac{17}{6}$ and remainder $-17$.

So $$\frac{s^2 + 18s}{6s + 6} = \frac{s}{6} + \frac{17}{6} - \frac{17}{6s + 6}$$

 

[sotownsend]

Answer:

The question wanted the integration of the final term of the derivation of the chain rule.

So...

∫[(s^2 + 18s)/(6s+6)] =

∫(s^2 / (6s + 6) + ∫(18s / (6s + 6)

The integral of these two terms yields a s^2 term with -1/12 as the coefficient. (Takes a lot of trig transforms / log transforms however)

 

[Stephen Tashi]

$$-\int (\frac{s}{6} + \frac{17}{6} - \frac{17}{6s + 6}) ds = \frac{-1}{12}s^2 - \frac{17}{6}s + \frac{17}{6} \log(6x + 6)$$