[QUESTION] integration (1 Viewer)

[richz]

hello

can some one help me with this q,

Integrate x^2/(4x^2 -1)

Hint: use partial fractions.

thnx

 

[FinalFantasy]

hello

can some one help me with this q,

Integrate x^2/(4x^2 -1)

Hint: use partial fractions.

thnx

x^2/(4x^2 -1)=(1\4)+(1\4)(4x²-1)
so integrate it is x\4+(1\4)int. 1\(4x²-1) dx
let 1\(4x²-1)=A\(2x+1)+B\(2x-1)
1=A(2x-1)+B(2x+1)
put x=1\2, .: 1=2B, B=1\2
put=-1\2, .: 1=-2A, A=-1\2
int. 1\(4x²-1) dx=-(1\4)ln |2x+1|+(1\4)ln|2x-1|

 

[Slidey]

hello

can some one help me with this q,

Integrate x^2/(4x^2 -1)

Hint: use partial fractions.

thnx

You can only use the partial fraction method when the degree of the numerator is less than the degree of the denominator.

So to meet this condition, perform polynomial division:
x^2 divided by 4x^2 -1

 

[withoutaface]

You can only use the partial fraction method when the degree of the numerator is less than the degree of the denominator.

So to meet this condition, perform polynomial division:
x^2 divided by 4x^2 -1

The degree of the numerator is equal to the denominator, so we can just split it up like so:
(x^2-1/4)/(4x^2-1)+1/4, if I remember my terms correctly, that's partial fractions.

 

[richz]

thnx guys