Integral of a quotient

**Glitch** · October 15th 2010, 02:05 AM

The question:
Find the integral of $\int^9_4{\frac{x^3 - x}{x^{\frac{3}{2}}}}$

I'm not sure where to start. My first thought was to split it into two integrals like so:

$\int^9_4(x^3 - x) \int^9_4{\frac{1}{x^{\frac{3}{2}}}}$

But from what I can tell, you cannot split an integral like that. Any ideas? Thanks.

**mr fantastic** · October 15th 2010, 02:11 AM

Originally Posted by Glitch

The question:
Find the integral of $\int^9_4{\frac{x^3 - x}{x^{\frac{3}{2}}}}$

I'm not sure where to start. My first thought was to split it into two integrals like so:

$\int^9_4(x^3 - x) \int^9_4{\frac{1}{x^{\frac{3}{2}}}}$

But from what I can tell, you cannot split an integral like that. Any ideas? Thanks.

Sorry, but what you have done is very wrong. Would you find $\displaystyle \int x^2 \, dx$ by splitting it up as $\displaystyle \int x \, dx \int x \, dx$ ?

You are expected to note that $\displaystyle \frac{x^3 - x}{x^{3/2}} = \frac{x^2 - 1}{x^{1/2}} = \frac{x^2}{x^{1/2}} - \frac{1}{x^{1/2}} = x^{3/2} - x^{-1/2}$ .

**Glitch** · October 15th 2010, 02:23 AM

Ahh yes, I realised I could do that not long after posting. >_<

Just to be sure, does this look correct?

$^9_4[\frac{2}{7}x^{\frac{7}{2}}] - ^9_4[-2x^{\frac{1}{2}}]$

**mr fantastic** · October 15th 2010, 02:28 AM

Originally Posted by Glitch

Ahh yes, I realised I could do that not long after posting. >_<

Just to be sure, does this look correct?

$^9_4[\frac{2}{7}x^{\frac{7}{2}}] - ^9_4[-2x^{\frac{1}{2}}]$

No. The first term is wrong and the second term contains an error.

It should be $^9_4[\frac{2}{5}x^{\frac{5}{2}}] - ^9_4[2x^{\frac{1}{2}}]$

**Glitch** · October 15th 2010, 02:31 AM

Wow, I really am going crazy. >_<

Thanks.

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