Integration Question (1 Viewer)

[rama_v]

Hi everyone:
Quick Question: How do you integrate

x^3/(2-x^2) ?

 

[Benny1103]

(x³)/(1-x²)
= -[(x³)/(x²-1)]
= -x[(x²)/(x²-1)]
= -x[(x²-1+1)/(x²-1)]
= -x(1+(1/(x²-1)))=
= -x -((x)/(x²-1))

Now just use partial fractions on the second one. Check my working but even if the working is incorrect, the basic idea is to add and subtract terms to 'simplify' the integrand.

 

[FinalFantasy]

wow, partial fractions?!!
isn't this 2unit?!

 

[rama_v]

Damn, this is a 3 unit question (I think harder 2 unit), I dont know partial fractions...It was in a 3 unit exam.

 

[FinalFantasy]

just use long division

 

[rama_v]

Do you mean divide x^3 by 2-x^2, as in polynomial division?
EDIT: never mind I got it thanks for the help.

 

[FinalFantasy]

x^3/(2-x^2) =2x\(2-x²)-x
so integrate u get -ln|2-x|-x²\2+C

 

[Benny1103]

Yeah your method is right Final Fantasy. I'm a bit rusty with integration and I wasn't aware that partial fractions isn't a part of the 2unit course.

 

[100percent]

Yeah your method is right Final Fantasy. I'm a bit rusty with integration and I wasn't aware that partial fractions isn't a part of the 2unit course.

partial fractions is not even 3u