Question

Hello.

I can't solve this:
and I might need a lot of help till I get this, since I did last time in school only with A and B but not with A BX and C ....

\[\begin{array}{l}\int {\frac{{xdx}}{{\left( {x - 1} \right)\left( {{x^2} + 1} \right)}}} \\\\\frac{x}{{\left( {x - 1} \right)\left( {{x^2} + 1} \right)}} = \frac{A}{{\left( {x - 1} \right)}} + \frac{{BX + C}}{{{x^2} + 1}}\end{array}\]

Answer 1

\[A(x^{2} +1) + (Bx+C)(x-1)\]


solve find A,B,C

Answer 2

where is equal sign? O_O

Answer 3

just take the LCM and u get x^2(A+B)+x(C-B)+A-C compare this with x and find A,Band C

Answer 4

EQUAL TO X .. SORRY...

Answer 5

alright so do common denominator on the right side and you will get A(x^2+1) + (x-1)(Bx+c) = x since the denominator cancels out,

then you got Ax^2 + A +Bx^2 +cx -bx -c

put the like terms together so you got (A+B)x^2 + (C-B)x + (A-c) = x

now since on the right side we only have we know A+B =0, C-B=1 A-C = 0
so now solve these 3 equations

Answer 6

A+B=0 since the coefficient of x^2 on the left hand side is zero
proceed in similar way,make equations and solve for A,B and C

Answer 7

lust give me a second, I'm lost :$

Answer 8

so we got c = 1+b then a-1-b=0 and we know a=-b solve that in we get -b-1-b=0 and then -2b=1 and b=-1/2 and a = 1/2 and then c = 1/2 :)

Answer 9

Thank you to everyone, I had terrible problems forming a system lol, I'll collect my brain now :A ... Thanks a lot ;)

Answer 10

@hamza_b23 thank you so much :)

Answer 11

noo problem goodluck :)