integral (1 Viewer)

nightweaver066 · Apr 23, 2012

Unsure if you'll need to take absolute value of the denominator but..

$\int \frac{x^3 - 2}{x^3 + 1}dx$

$= \int 1 - \frac{3}{x^3 + 1}dx$

$= x - \int \frac{3}{(x + 1)(x^2 - x + 1)}$

$\frac{3}{(x + 1)(x^2 - x + 1)} \equiv \frac{A}{x + 1} + \frac{Bx + C}{x^2 - x + 1}$

$And go from there$

Timothy.Siu · Apr 23, 2012

The absolute value does matter.
It'll just mean you have cases for the final answer, if x<-1 or x>=-1

juantheron · Apr 23, 2012

Thanks friends for answering.

I soorry for my typo mistake,actually original question is $\displaystyle \int \frac{x^3-2}{\sqrt{(x^3+1)^3}}dx$

nightweaver066 · Apr 23, 2012

Timothy.Siu said:
The absolute value does matter.
It'll just mean you have cases for the final answer, if x<-1 or x>=-1

Okay thanks

barbernator · Apr 23, 2012

nightweaver066 said:
Use substitution, let u = x^3 + 1

does this work? i can't get it to go

nightweaver066 · Apr 23, 2012

barbernator said:
does this work? i can't get it to go

I just tried it myself and it didn't work lol.

Would attempt it again but i'm busy at the moment..

barbernator · Apr 23, 2012

nightweaver066 said:
I just tried it myself and it didn't work lol.

Would attempt it again but i'm busy at the moment..

thats fine, ill find a way!

deswa1 · Apr 23, 2012

barbernator said:
thats fine, ill find a way!

I don't think you can do it with substitution or by parts or any 'normal' means. The way I did it was trial and error- working out what you need to differentiate to give that as the answer. I can post the answer though if you guys want (but that might spoil it...)

barbernator · Apr 23, 2012

deswa1 said:
I don't think you can do it with substitution or by parts or any 'normal' means. The way I did it was trial and error- working out what you need to differentiate to give that as the answer. I can post the answer though if you guys want (but that might spoil it...)

hmmm interesting. Why would they ask a question about it then? wolfram alpha didnt even want to do the steps for me. It said its unavailable for some reason. But thats pretty cool how you found it by trial and error.

deswa1 · Apr 23, 2012

barbernator said:
hmmm interesting. Why would they ask a question about it then? wolfram alpha didnt even want to do the steps for me. It said its unavailable for some reason. But thats pretty cool how you found it by trial and error.

I don't think they would- I've never seen anything this hard in any HSC paper (the fact that I got it by trial was a fluke pretty much). Where did you get this question from OP?

juantheron · Apr 25, 2012

CcYann · Apr 25, 2012

juantheron said:
$\hspace{-16}\bf{\int\frac{x^3-2}{\sqrt{(x^3+1)^3}}dx}$\\\\\\ Using Hit and Trail......\\\\ Let $\bf{\frac{d}{dx}\left\{\frac{Ax+B}{\sqrt{x^3+1}}\right\}=\frac{x^3-2}{\sqrt{(x^3+1)^3}}}$\\\\\\ $\bf{\frac{2(x^3+1).A-(Ax+B).3x^2}{\sqrt{(x^3+1)^3}}=\frac{2x^3-4}{\sqrt{(x^3+1)^3}}}$\\\\\\ $\bf{\frac{-Ax^3+2A-3Bx^2}{\sqrt{(x^3+1)^3}}=\frac{2x^3-4}{\sqrt{(x^3+1)^3}}}$\\\\\\ Now Camparing, We Get\\\\ $\bf{A=-2\;\;,B=0}$\\\\\\ So $\bf{\frac{d}{dx}\left\{\frac{-x+0}{\sqrt{x^3+1}}\right\}=\frac{x^3-2}{\sqrt{(x^3+1)^3}}}$\\\\\\ Now Integrate Both Side, We Get\\\\ $\bf{\int \frac{d}{dx}\left\{\frac{-2x}{\sqrt{x^3+1}}\right\}dx=\int \frac{x^3-2}{\sqrt{(x^3+1)^3}}dx}$\\\\\\ So $\bf{\int \frac{x^3-2}{\sqrt{(x^3+1)^3}}dx=\frac{-2x}{\sqrt{x^3+1}}+C}$$

nice

CcYann · Apr 25, 2012

I don't understand these steps tho. can somebody explain for me plz?

integral (1 Viewer)

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