can someone help me solve this with partial fractions? I'll

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here's another one :\ hopefully someone will help me next morning

(As long as the numerators degree is less than or equal to the denominators)
(If not use long division -See Q 10)

The point of partial fractions are to split a difficult fraction into smaller parts that are easier to integrate.

First simplify the denominator. (Factorise)
If you see (x-a), one component will be A/(x-a)
If you see (x+b)^n, there will be multiple components A/(x+b)^1 + B/(x+b)^2 + ...+ C/(x+b)^(n-1) + D/(x+b)^n
If you see (x^2 + cx +d) or any combination with an x^2, one component will be (Ax + B)/(x^2 + cx +d)

In general the partial fraction will have a degree less than the denominator.
Once you have all the partial fractions you can solve for the unknowns.
Integrate each function individually.

Questions Next

Q9

(answer you provided is wrong) that would be the correct answer for (x^2)/[(x-1)(x+1)^2]

Can be split up to A/(x-1) + (Bx + C)/(x^2+1)
Combine fractions [A(x^2+1) + (Bx+C)(x-1)]/[(x-1)(x^2+1)] is the same as your original fraction.
Thus numerators are the same: x^2 = A(x^2+1) + (Bx+C)(x-1)

Solving for A,B,C can be tedious, but there are several methods. You can equate coefficients.
(x^2): 1 = A + B
(x): 0 = C - B
(0): 0 = A - C

OR sub in values

when x = 1;
(1)^2 = A((1)^2+1) + (Bx + C)((1)-1)
1 = 2A
A = 0.5
B = 0.5
C = 0.5

Once you have the fractions 1/[2(x-1)] + (0.5x + 0.5)/(x^2+1) use simple integration.

The second fraction is kind of difficult so we split it up to be simpler.
x/[2(x^2+1)] and 1/[2(x^2+1)]

The first is simple because the derivative is on the numerator.
The second is an arcTan (Tan^-1)

Thus final answer:
ln(x^2+1)/4 + ln(x-1)/2 + tan-1(x)/2

Forgot a + C at the end of Q9

Q10 is a little more difficult because it the numerators degree is larger than the denominators.

Again I think the answers are wrong.

So we use long division.
"how many times does x^2 go into x^3" - x times.
The muliply the x^2 bit by x to give:
x^3 -5x^2 +6x.
then subtract it from x^3 -1
(keep going)
_____________________x+5
x^2 -5x +6 ) x^3 + 0x^2 + 0x -1
________-_x^3 -5x^2 +6x
_________ 0 + 5x^2 -6x -1
___________-_5x^2 -25x +30
_____________0x^2 +19x -31
R = 19x -31

Thus the integral = x+5 + (19x - 31)/(x^2 -5x+6)
Then we do partial fractions on the remaining bit.

As (x^2 -5x+6) = (x-2)(x-3)

(19x - 31)/[(x-2)(x-3)] = A/(x-2) + B/(x-3)

19x - 31 = A(x-3) + B(x-2)
Equating coefficients:
(x): 19 = A + B
(0) -31 = -3A -2B

A = 26
B = -7

Thus integral = x + 5 - 7/(x-2) + 26/(x-3)
Answer = x^2/2 + 5x -7ln(x-2) + 26ln(x-3) + C

A good thing to do is to check your answers for the patial fraction bit using wolfram alpha.

Eg: http://www.wolframalpha.com/input/?i=partial+fractions+(x%5E3-1)%2F(x%5E2-5x%2B6)

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thank you