i need help with some integration...

Question

anonymous · Answer

$$(6x-9)/(x^2-3x+5)^3$$
i know i need to let u=x^2-3x+5  
and thus du = (2x-3)dx 
         so, 3du = (6x-9)dx

now here is where things start to get hazy lol please explain without completely giving it to me. 

Thank you very much

anonymous · Answer

the next step i went to was 
$$3\int\limits_{}^{} 1/u^3 $$

is that right?

anonymous · Answer

$$3\int\limits {\frac{du}{u^2}}$$

anonymous · Answer

why u^2?

anonymous · Answer

Yes.  
$$3\int\limits\limits_{}^{} \frac{1}{u^3}du$$

anonymous · Answer

sweet! so now i integrate and i get

$$(-1/4)(3)(\ln(u^-4))$$

anonymous · Answer

What ??

anonymous · Answer

wait haha i think i see my mistake

$$(3)(-1/4)(1/u^4)$$

anonymous · Answer

It is a indefinite integral, you cannot have a real number

anonymous · Answer

$$\frac{-3}{2} \frac{1}{u^2} +c$$

anonymous · Answer

oh.. ok i think i get that

anonymous · Answer

no no shivam_bhalla

anonymous · Answer

Remember,
$$\int\limits_{}^{}u^ndu = \frac{u^{n+1}}{n+1} +c$$

anonymous · Answer

yup, but $$u^n=u^-2$$

anonymous · Answer

@anhkhoavo1210 ,  I think you need some sleep :P

anonymous · Answer

right, the negative got me mixed up, but my answer now is

$$-3/2(1/(x^2-3x+5))$$

anonymous · Answer

@m.auld64 , You are messing up with power

anonymous · Answer

your above result is $$\frac{-3}{2u^2}+C$$

anonymous · Answer

oops add a ^2 on the end of the denominator lol

anonymous · Answer

That's better :)

anonymous · Answer

@anhkhoavo1210 , What's the problem??  n=-3
So 
$$3 \int\limits_{}^{}u^{-3}du = 3 \frac{u^{-3+1}}{-3+1} +c$$
Any problem with that ???

anonymous · Answer

Sorry, i see 3 become 2 :)

anonymous · Answer

LOL. Next time write it down and try :D