Тема: Find partial fraction decomposition of (2x)/(x^3+x^2)?

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(6x² - 3x + 1) / [(4x + 1)(x² + 1) = A / (4x + 1) + (Bx + C) / (x² + 1) 6x² - 3x + 1 = A(x² + 1) + (Bx + C)(4x + 1) You would be correct to work on 4x + 1 = 0 (which means x = -1/4). This will enable you to work out the value of A. Because of the problem you outlined we now proceed as follows. Let x = 0 (because x = 0 is the simplest case) and this will give us a value for C. (We also need touse our previously calculated value for A) Then let x = 1(the next simplest value to use) and this will give us a value for B. (We also need to use our previously calculated values for A and C).

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The equation for partial pressure is: Pi=XiPt Pi=partial pressure Xi=mol fraction Pt=total pressure Okay, so we are given grams, so the first thing we need to do, is convert them into mols: (32g O2) x (1mol O2/32 g O2) = 1 mol O2 (45 g N2) x (1 mol N2/28 g N2)=1.61 mol N2 (12 g C) x (1 mol C/12 g C)=1. mol C Now, we need to calculate the mol fraction which is: (mol of component)/(total mols) so we want: (1 mol C + 1 mol O2)/ (1 mol O2 +1.61 mol N2 + 1 mol C) and we get: 0.554 (there is no unit, because mol/mol cancels out) So now to do partial pressure: P(CO2)=X(CO2)Pt P(CO2)=(0.554)(4.2 atm) P(CO2)=2.33 atm

1/(4x^2 - 9) can be rewritten as 1/[(2x + 3)(2x - 3)]. In which case, you should get something that looks like this: 1 = A(2x - 3) + B(2x + 3) Solving for A and B I get: A = -1/6 B = 1/6 So the new function looks like: (1/6)/(2x - 3) - (1/6)/(2x + 3), in which case the integral would look like: (1/12)ln(2x - 3) - (1/12)ln(2x + 3) + C [Technically, those parentheses are supposed to be absolute value signs]


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