Solving Partial Fractions with Polynomial Division

beaf123
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Homework Statement



∫ (x^3)/(x^2+2x+1)

I think I could solve it if I knew how they did this operation:

From the solution:
'
(x^3)/(x^2+2x+1) = (x-2) + (3x+2)/(x+1)^2 ( After long division)

Did they use polynomialdivision?

x^3: x^2-2X+1=

If so, how?
 
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beaf123 said:

Homework Statement



∫ (x^3)/(x^2+2x+1) dx

I think I could solve it if I knew how they did this operation:

From the solution:
'
(x^3)/(x^2+2x+1) = (x-2) + (3x+2)/(x+1)^2 ( After long division)

Did they use polynomial division?

x^3: x^2-2X+1=

If so, how?

Yes, they used long division for polynomials.

Here's a link to Wikipedia's page on polynomial long division.

If you don't want to use long division, use the substitution u = x+1.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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