Partial Fractions Q7 (1 Viewer)

[want2beSMART]

INTEGRAL of x/(x^4-1)dx

 

[FinalFantasy]

INTEGRAL of x/(x^4-1)dx

x\(x^4-1)=x\(x²+1)(x²-1)=x\(x²+1)(x+1)(x-1)
let x\(x²+1)(x+1)(x-1)=(Ax+B)\(x²+1)+C\(x+1)+D\(x-1)
x=(Ax+B)(x²-1)+C(x²+1)(x-1)+D(x²+1)(x+1)
put=1, therefore 1=D(2)(2)=4D, therefore D=1\4
put x=-1, therefore -1=-4C, .: C=1\4
put x=0, .: 0=-B-1\4+1\4, therefore B=0
put x=2, .: 2=6A+1\4(5)+1\4(5)(3), .: A=-1\2

 

[want2beSMART]

erm why put x=2?

have you completed the question?

 

[Slidey]

Couldn't you just go
x=(Ax+B)(x^2-1)+(Cx+D)(x^2+1)
And get:
Integral = (ln(x^2+1)-ln(x^2-1))/4
=ln((x^2+1)/(x^2-1))/4 + C

 

[FinalFantasy]

i didn't complete the q. i just did the partial fraction bit

 

[want2beSMART]

na the answer is:

1/4 ln|x-1| + 1/4 ln|x+1| - 1/4 ln|x^2 +1| + C

 

[want2beSMART]

i didn't complete the q. i just did the partial fraction bit

finalfantasy, could you complete the question?

i got A = -1/4 and B = 1/4

i made A/ (x+1) + B/ (x-1) + (Cx + D)/(x^2 + 1)

 

[FinalFantasy]

INTEGRAL of x/(x^4-1)dx

x\(x^4-1)=x\(x²+1)(x²-1)=x\(x²+1)(x+1)(x-1)
let x\(x²+1)(x+1)(x-1)=(Ax+B)\(x²+1)+C\(x+1)+D\(x-1)
x=(Ax+B)(x²-1)+C(x²+1)(x-1)+D(x²+1)(x+1)
put=1, therefore 1=D(2)(2)=4D, therefore D=1\4
put x=-1, therefore -1=-4C, .: C=1\4
put x=0, .: 0=-B-1\4+1\4, therefore B=0
put x=2, .: 2=6A+1\4(5)+1\4(5)(3), .: A=-1\2

therefore Int. x\(x^4-1) dx=-1\2int. x\(x²+1)dx+1\4int.1\(x+1)dx+1\4int. 1\(x-1)dx
=-1\4ln|x²+1|+1\4ln|x+1|+1\4ln|x-1|+C

 

[Slidey]

na the answer is:

1/4 ln|x-1| + 1/4 ln|x+1| - 1/4 ln|x^2 +1| + C

You DO realise that is the same as my answer... right?

 

[FinalFantasy]

lolol.............................

 

[want2beSMART]

nope cbb realising