integral(5(x^4)+4(x^5))/(((x^5+x+1))^2)

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- Jun 6th 2011, 11:35 AM #1

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- Jun 6th 2011, 12:28 PM #2

- Jun 6th 2011, 01:09 PM #3

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- Jun 6th 2011, 01:49 PM #4

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Bit of give and take in the numerator...

$\displaystyle \int \frac{5x^4 + 4x^5}{(x^5 + x + 1)^2}$

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

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

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

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

Do parts on

$\displaystyle =\ \int \frac{(5x^4 + 1)}{(x^5 + x + 1)^2} (1 + x)$

(integrating the fraction)...

Edit: Just in case a picture helps...

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- Jun 6th 2011, 02:42 PM #5

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What a strange integral... considering the amount of trouble you have to go through to solve it (using any "normal" method) you'd think this can't have a nice solution but it has (I used Mathematica - sorry if that feels for you like I cheated) and wound up with something that was... well, funny how "nice" it was.

Anyway, all you have to do is note that the form tom@ballooncalculus wrote can be rewritten as:

$\displaystyle \int \frac{(5x^4+1)(x+1)-(x^5+x+1)}{(x^5+x+1)^2}\,dx$

$\displaystyle =-\int \frac{(x+1)'(x^5+x+1)-(x+1)(x^5+x+1)'}{(x^5+x+1)^2}\,dx$

And that is practically the definition of the quotient rule, so:

$\displaystyle =-\int(\frac{x+1}{x^5+x+1})'\,dx=-\frac{x+1}{x^5+x+1}$

Of course you get the same result by substituting

$\displaystyle t=\frac{x+1}{x^5+x+1}$.

Now, since any other way I can think of would require evaluating

$\displaystyle \int\frac{\,dx}{x^5+x+1}$, I guess this is the way it's supposed to be solved. I also congratulate anyone who would have been able to do that .

- Jun 6th 2011, 04:37 PM #6

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one might guess, because the numerator is a square, that the integral may be some quotient.

differentiating $\displaystyle \frac{f(x)}{x^5+x+1}$ gives:

$\displaystyle \frac{(x^5+x+1)f'(x) - f(x)(5x^4 + 1)}{(x^5+x+1)^2}$, which we want to be equal to $\displaystyle \frac{4x^5+5x^4}{(x^5+x+1)^2}$

that is, we want $\displaystyle (x^5+x+1)f'(x) - (5x^4 + 1)f(x) = 4x^5 + 5x^4 $.

if f(x) is a polynomial, with leading term of degree n:

$\displaystyle f(x) = ax^n + r(x), f'(x) = nax^{n-1}+r'(x)$, then the leading term of

$\displaystyle (x^5+x+1)f'(x) - (5x^4 + 1)f(x)$ is $\displaystyle a(n-5)x^{n+4}$.

this tells us n = 1, and a = -1, that is f(x) = -x + b, for some real number b.

then $\displaystyle (x^5+x+1)f'(x) - (5x^4 + 1)f(x) = -x^5-x-1+5x^5+x-5bx^4-b$

$\displaystyle =4x^5-5bx^4-(b+1) = 4x^5+5x^4$, which only holds when b = -1.

we can thus conclude that f(x) = -x - 1