Evaluate the following integrals

http://www.cramster.com/Answer-Board...4200008485.gif

http://www.cramster.com/Answer-Board...5545000747.gif

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- May 16th 2008, 07:31 AMmatty888Easy Integrals!!
Evaluate the following integrals

http://www.cramster.com/Answer-Board...4200008485.gif

http://www.cramster.com/Answer-Board...5545000747.gif - May 16th 2008, 07:42 AMDr Zoidburg
1st one comes out to:

$\displaystyle \frac{1}{4}x^2 (2log(x) - 1)$

2nd one comes out to:

$\displaystyle \frac{1}{5}(2log(x - 2) + 3log(x + 3))$

I think! not 100% certain there. - May 16th 2008, 07:53 AMTheEmptySet
- May 16th 2008, 08:04 AMflyingsquirrel
Hi

Let's find the partial fraction decomposition of $\displaystyle \frac{x}{x^2+x-6}$.

The discriminant of the denominator is $\displaystyle \Delta = 1^2-4\cdot1\cdot (-6)=25>0$ hence the denominator has to different real roots : $\displaystyle \frac{-1\pm\sqrt{25}}{2}$ that is to say -3 and 2. Hence $\displaystyle x^2+x-6=(x+3)(x-2)$

It gives $\displaystyle

\frac{x}{x^2+x-6} = \frac{x}{(x+3)(x-2)}=\frac{\alpha}{x+3}+\frac{\beta}{x-2}$ which has to be solved for $\displaystyle \alpha$ and $\displaystyle \beta$.

Can you take it from here ? - May 16th 2008, 12:58 PMKrizalid
$\displaystyle \begin{aligned}

\frac{x}{x^{2}+x-6}&=\frac{2x+1-1}{2(x-2)(x+3)} \\

& =\frac{1}{2}\left[ \frac{2x+1}{x^{2}+x-6}-\frac{(x+3)-(x-2)}{5(x-2)(x+3)} \right] \\

& =\frac{1}{2}\left[ \frac{2x+1}{x^{2}+x-6}-\frac{1}{5}\left\{ \frac{1}{x-2}-\frac{1}{x+3} \right\} \right].

\end{aligned}$

Integrate.

P.S.: first integral can be tackled by double integration technique too. (Sun) - May 16th 2008, 03:43 PMJhevon