Easy Integrals!!

**matty888** · May 16th 2008, 07:31 AM

Evaluate the following integrals

**Dr Zoidburg** · May 16th 2008, 07:42 AM

1st one comes out to:
$\frac{1}{4}x^2 (2log(x) - 1)$

2nd one comes out to:
$\frac{1}{5}(2log(x - 2) + 3log(x + 3))$

I think! not 100% certain there.

**TheEmptySet** · May 16th 2008, 07:53 AM

Originally Posted by matty888

Evaluate the following integrals

for the first one use the change of base formula to get

$\int_{1}^{2}x\log(x)dx=\int_{1}^{2}x\frac{\ln(x)}{ \ln(10)}dx=\frac{1}{\ln(10)}\int_{1}^{2}x\ln(x)dx$

use integration by parts with u=ln(x) dv=x

That should get you started

**flyingsquirrel** · May 16th 2008, 08:04 AM

Hi

Originally Posted by matty888

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

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

It gives $ \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 $\alpha$ and $\beta$ .

Can you take it from here ?

**Krizalid** · May 16th 2008, 12:58 PM

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

Integrate.

P.S.: first integral can be tackled by double integration technique too.

**Jhevon** · May 16th 2008, 03:43 PM

Originally Posted by Krizalid

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

Integrate.

Haha! i LOVE seeing when you do that!

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