Partial Fraction Integration Help =(

[hellos]

can anyone help me with this integration

integral (dx)/(x^3+x)

i tried solving it with partial integration and got stuck when i got find B and C...this was what i did

= integral (dx)/[x(x^2 + 1)]

= 1/[x(x^2 - 1)] = A/x + (Bx+C)/(x^2 + 1)

=> 1 = A(x^2 + 1) + (Bx+C)(x)

=> 1 = Ax^2 +A + Bx^2 + Cx

A = 1

then i got stuck here on how to get the B and C variable

Thanks in advance to those who can help

 

[neutrino]

The question says [tex]\int\frac{dx}{x^3+x}[/tex] while your answer contains [tex]\int\frac{dx}{x(x^2-1)}[/tex]. Which is right?

 

[HallsofIvy]

can anyone help me with this integration

integral (dx)/(x^3+x)

i tried solving it with partial integration and got stuck when i got find B and C...this was what i did

= integral (dx)/[x(x^2 - 1)]

= 1/[x(x^2 - 1)] = A/x + (Bx+C)/(x^2 - 1)

=> 1 = A(x^2 - 1) + (Bx+C)(x)

=> 1 = Ax^2 - A + Bx^2 + Cx

A = -1

then i got stuck here on how to get the B and C variable

Thanks in advance to those who can help

?Is your integrand [tex]\frac{1}{x^3+ x}[/tex] or [tex]\frac{1}{x^3- x}[/tex]? You give both.

If it is [tex]\frac{1}{x^3-x}[/tex] then you need to factor more:
x³- x= x(x²-1)= x(x-1)(x+1) so
[tex]\frac{1}{x^3- x}= \frac{A}{x}+ \frac{B}{x-1}+ \frac{C}{x+1}[/tex]
Then 1= A(x-1)(x+1)+ B(x)(x+1)+ C(x)(x-1) and you can determine A, B, C by letting x= -1, 0, 1 in turn.

If it is [itex]\frac{1}{x^3+ x}[/itex] then it is a little harder:
x³+ x= x(x²+ 1) and x²+ 1 cannot be factored further (in terms of real numbers).
[tex]\frac{1}{x^3+x}= \frac{A}{x}+ \frac{Bx+C}{x^2+ 1}[/tex]
then 1= A(x²+1)+ (Bx+ C)x.

Taking x= 0, A= 1. Now there are 3 different ways to determine B and C.

a) Subtract 1/x from the fraction:
[tex]\frac{1}{x^3+x}- \frac{1}{x}= \frac{1}{x^3+x}- \frac{x^2+1}{x^3+x}= -\frac{x}{x^2+1}[/tex]
so that the Bx+ C in [itex]\frac{Bx+C}{x^2+1}[/itex] is clearly B= -1, C= 0.

b) Compare like powers of x:
with 1= A(x²+1)+ (Bx+ C)x and A= 1, we have 1= x²+ 1+ (Bx+ C)x= (B+1)x^2+ Cx+ 1. We must have B+1= x and C= 0 giving, again, B= -1, C= 0.

c) Use some values of x other than 0:
1= 1(x²+1)+ (Bx+ C)x.
Taking x= 1, 1= 2+ B+C and taking x= -1, 1= 2+(-B+C)(-1)= 2+B-C
Now solve the two equation B+ C= -1 and B-C= -1 to get, once again,
B= -1, C= 0.

 

[hellos]

Thanks for the input guys