# Integration Challenge

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• Jun 6th 2006, 05:54 AM
nath_quam
Integration Challenge
Prove the integral of 1/(x^3+8) between the x terminals 4 and 1 equals (pie root 3)/36
• Jun 6th 2006, 01:08 PM
ThePerfectHacker
Quote:

Originally Posted by nath_quam
Prove the integral of 1/(x^3+8) between the x terminals 4 and 1 equals (pie root 3)/36

You can express,
$\frac{1}{x^3+8}$
As,
$\frac{1}{(x+2)(x^2-2x+4)}$
Then decompose this fraction as,
$\frac{A}{x+2}+\frac{Bx+C}{x^2-2x+4}$

When fully decomposed gives,
$
\frac{2}{3}\cdot \int^4_1 \frac{1/8}{x+2}+\frac{-1/8x+1/2}{x^2-2x+4} dx$

The first summand calculates to,
$\left \int^4_1 \frac{1/8}{x+2}dx=\ln |x+2| \right|^4_1=\ln 6-\ln 3=\ln 2$
Multiply by constant 1/8 and 2/3,
$(1/12)\ln 2$.

The second summand need manipulation,
$\int^4_1 \frac{-1/8x+1/2}{x^2-2x+4} dx=-\frac{1}{8}\int^4_1 \frac{(x-1)-3}{(x-1)^2+3} dx$

Apply a linear substitution, $u=x-1$
$\int^3_0 \frac{u-3}{u^2+3}du$
Express as,
$\int^3_0 \frac{u}{u^2+3}-\frac{1}{(\frac{u}{\sqrt{3}})^2+1} du$

Yields, (first summand is natural logarithm in disuise, the second is the arctanget),
$\left \frac{1}{2}\ln|u^2+3| \right|^3_0 - \sqrt{3} \left \tan^{-1} \left( \frac{u}{\sqrt{3}} \right) \right|^3_0$
The first summand gives,
$\frac{1}{2}\ln 12-\frac{1}{2}\ln 3=\frac{1}{2} (\ln 12-\ln 3)=\frac{1}{2}\ln 4=\ln 4^{1/2}=\ln 2$
The second summand gives,
$\sqrt{3} \tan^{-1} \left( \frac{3}{\sqrt{3}} \right) -\sqrt{3} \tan^{-1} (0)=\frac{\pi\sqrt{3}}{3}$
The next step is to subtract these summands,
$\ln 2-\frac{\pi\sqrt{3}}{3}$
Next is to mutiply this by -1/8 thus,
$-(1/8)\ln 2+\frac{\pi\sqrt{3}}{24}$
Now mutiply to 2/3,
$-(1/12)\ln 2+\frac{\pi\sqrt{3}}{36}$
And add your very first summand from the beginning,
thus,
$(1/12)\ln 2-(1/12)\ln 2+\frac{\pi\sqrt{3}}{36}=\frac{\pi\sqrt{3}}{36}$

$\mathbb{Q}.\mathbb{E}.\mathbb{D}$