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Partial Fractions

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


x^2+x = A(x^2 + 1) + B(x^2 - 1)

Let x = 1

A=1

How do I find B?

Thanks
Reply 1
Original post by khanpatel321
x^2 + x / (x^2 - 1)(x^2 + 1)


x^2+x = A(x^2 + 1) + B(x^2 - 1)

Let x = 1

A=1

How do I find B?

Thanks


x2 - 1 factorizes further
x2+1 is irreducible
Original post by khanpatel321
x^2 + x / (x^2 - 1)(x^2 + 1)


x^2+x = A(x^2 + 1) + B(x^2 - 1)

Let x = 1

A=1

How do I find B?

Thanks


You want Ax+Bx2+1+Cx+1+Dx1\frac{Ax+B}{x^2+1}+\frac{C}{x+1}+\frac{D}{x-1}
Original post by TeeEm
x2 - 1 factorizes further
x2+1 is irreducible


Original post by brianeverit
You want Ax+Bx2+1+Cx+1+Dx1\frac{Ax+B}{x^2+1}+\frac{C}{x+1}+\frac{D}{x-1}


I thought you only add a polynomial when the fraction is an improper fraction?

Why isn't it

A/(x^2 + 1) + B/(x-1) + C/(x+1)
Reply 4
Original post by khanpatel321
I thought you only add a polynomial when the fraction is an improper fraction?

Why isn't it

A/(x^2 + 1) + B/(x-1) + C/(x+1)


brianeverit has given you the correct form ...

the reason for what you are asking you may not understand at present but here it comes...
when you divide by an irreducible quadratic the most general form for the remainder is a linear expression
Original post by TeeEm
brianeverit has given you the correct form ...

the reason for what you are asking you may not understand at present but here it comes...
when you divide by an irreducible quadratic the most general form for the remainder is a linear expression


Original post by khanpatel321
I thought you only add a polynomial when the fraction is an improper fraction?

Why isn't it

A/(x^2 + 1) + B/(x-1) + C/(x+1)


Original post by brianeverit
You want Ax+Bx2+1+Cx+1+Dx1\frac{Ax+B}{x^2+1}+\frac{C}{x+1}+\frac{D}{x-1}



Ok I understand now. Just one last question... when differentiating sin(2x) the answer is 2cos(2x) but when you differentiate 2sin(-2x) the answer is -2cos(2x) why isn't it -2cos(-2x) ?
Reply 6
Original post by khanpatel321
Ok I understand now. Just one last question... when differentiating sin(2x) the answer is 2cos(2x) but when you differentiate 2sin(-2x) the answer is -2cos(2x) why isn't it -2cos(-2x) ?


cos is even

cos(2x) is the same as cos(-2x)
Original post by TeeEm
cos is even

cos(2x) is the same as cos(-2x)


e^(-x) sin(-2x)

d/dx = -e^(-x)sin(-2x) - 2e(^-x)cos(-2x)

is this correct?
Reply 8
Original post by khanpatel321
e^(-x) sin(-2x)

d/dx = -e^(-x)sin(-2x) - 2e(^-x)cos(-2x)

is this correct?


It is but I would have written

d/dx = e^(-x)sin(2x) - 2e(^-x)cos(2x)
Reply 9
Original post by khanpatel321
Ok I understand now. Just one last question... when differentiating sin(2x) the answer is 2cos(2x) but when you differentiate 2sin(-2x) the answer is -2cos(2x) why isn't it -2cos(-2x) ?


ddx[2sin(2x)]=4cos(2x)\displaystyle \frac{\mathrm{d}}{\mathrm{d}x} \bigg[2 \sin (-2x)\bigg] = -4 \cos (2x)

Since cosine is an even function. That is f(x)=cosxf(x) = \cos x has the property that f(x)=f(x)f(x) = f(-x).

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