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HALPPPP! IM DYING!!!!!!!!!1

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Q)1 ((x^3)+1) / (((x^2) -2)^2)
that meas x cube plus one in the numerator... and the square of x square minus 2 in the denominator
 
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Q)2 x cube minus two x square plus three x minus four ... in the numerator
and x square plus two x plus 2 multiplied by the whole square of x minus 1
lol
x^3 -2x^2 +3x -4
(x-1)^2 (x^2 +2x+2)
 
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ahem... now can someone do the partial fractions in this and make this in a form i can integrate?

x^2 -2x+3
(x-1)(x^2 +2x+2)

please do the partial fractions clearly...
Ever heard of the boy who cried wolf?
 
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ok..... srry for late reply...... but got bad news.... :(
1st one.
i really dont know what to do here... i mean, is it not already in partial fractions???
for 2nd one
OK i've never done such a partial fraction.... i'm not sure its good....... but i tried here......
i'll do the main work, u finish it all.....
by using the A, B, Cx + D thing u get this
A(x^2 + 2x + 2)(x-1) + B(x^2 + 2x + 2) + (Cx+D)((x-1)^2) for the numerator.....
now compare coefficients and all
u get all these equations!!!!
for x^3 ----> A+C=1
for x^2 -----> A+B+D-2C = -2
for x -------> 2B+C-2D = 3

now, u can eliminate x-1 by letting x =1
u will get B=-2/5

now if you let x=0, u get this equation
-2A+2B+D=-4
u replace B here.....
-2A+D = 24/5


now big task is solving AALLL these equationS!!!!!
ok firstly replace B in coeff of x equation.....
u get this C=2D+19/5
Replace this C in x^3 equation
u get an eqaution in terms of A and D
Now remember tht red equation as well is in terms of A and D
u solve these 2 and get the values of A and D
and the rest u just replace the values u get.....

Srry for late reply.... :(
 
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dun wrry yar... mindu the answrs are weird like 7/50 and 18/25 likethat...
the only reason i dunno how to do em is cux tese arennt in the a levels syllabus....
 
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ok..... srry for late reply...... but got bad news.... :(
1st one.
i really dont know what to do here... i mean, is it not already in partial fractions???
for 2nd one
OK i've never done such a partial fraction.... i'm not sure its good....... but i tried here......
i'll do the main work, u finish it all.....
by using the A, B, Cx + D thing u get this
A(x^2 + 2x + 2)(x-1) + B(x^2 + 2x + 2) + (Cx+D)((x-1)^2) for the numerator.....
now compare coefficients and all
u get all these equations!!!!
for x^3 ----> A+C=1
for x^2 -----> A+B+D-2C = -2
for x -------> 2B+C-2D = 3

now, u can eliminate x-1 by letting x =1
u will get B=-2/5

now if you let x=0, u get this equation
-2A+2B+D=-4
u replace B here.....
-2A+D = 24/5


now big task is solving AALLL these equationS!!!!!
ok firstly replace B in coeff of x equation.....
u get this C=2D+19/5
Replace this C in x^3 equation
u get an eqaution in terms of A and D
Now remember tht red equation as well is in terms of A and D
u solve these 2 and get the values of A and D
and the rest u just replace the values u get.....

Srry for late reply.... :(
atleast u helped :) thankuuuuuuuuuuuuuuuuuuuuuu
 
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If you don't mind me asking.. where are these questions from..? If I know where they are from.. I might be able to help you better...

((x^3)+1) / (((x^2) -2)^2)

That's already simplified... can't simplify it more..
 
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