Partial Fractions, what am I doing wrong? (Handwritten pic)

Well, it seems that I'm getting something wrong here, and I'm doing it for the second time now and keep getting same answers. Anyone care to help? Greatly appreciated :smile:

EDIT: I keep getting A=1, B=3, C=4 - but with these numbers part b is impossible to do as ln of a negative number appears

Reply 1

toronto353

19

Ok I would do the following:

You have determined the following:

15 - 17x = A( 1 - 3x)^2 + B(2 + x)(1 - 3x) + C(2 + x)

Let x = -2

therefore

49 = 49A

therefore A = 1

Now what you could do would be to compare coefficients here.

I suggest comparing coefficients on both sides of the equation of x^2

Therefore

0 = 9A - 3B

A = 1 therefore rearrange to get 9 = 3B therefore B = 3

Now compare coefficients of the constants.

Here

15= A + 2B + 2C

substitute your values of A and B in

15 = 1 + 6 + 2C

Therefore C = 4

Therefore A = 1, B = 3, C = 4

Sometimes comparing coefficients is better than using fractions

Reply 2

Moa

OP

13

toronto353

Ok I would do the following:

You have determined the following:

15 - 17x = A( 1 - 3x)^2 + B(2 + x)(1 - 3x) + C(2 + x)

Let x = -2

therefore

49 = 49A

therefore A = 1

Now what you could do would be to compare coefficients here.

I suggest comparing coefficients on both sides of the equation of x^2

Therefore

0 = 9A - 3B

A = 1 therefore rearrange to get 9 = 3B therefore B = 3

Now compare coefficients of the constants.

Here

15= A + 2B + 2C

substitute your values of A and B in

15 = 1 + 6 + 2C

Therefore C = 4

Therefore A = 1, B = 3, C = 4

Sometimes comparing coefficients is better than using fractions

This answer is apparently wrong, and this is same answer that I am gettng.

Check it on wolfram: http://www.wolframalpha.com/input/?i=partial+fractions+(15-17x)/((2%2Bx)(1-3x)^2)+

(Right answer is written in pen on the test paper)

Reply 3

DeanT

What paper is that? I get A = 1, B = 3, C = 4 too, and 1 + ln8 for the integration. (if the limits are [-1, 0])

Reply 4

Moa

OP

13

DeanT

What paper is that? I get A = 1, B = 3, C = 4 too, and 1 + ln8 for the integration. (if the limits are [-1, 0])

But limits are [0,-1], if I try to use them I get an error. Thats one of the Solomon Press papers.

Reply 5

DeanT

I did:

.

\displaystyle \int\frac{15 - 17x}{(2 + x)(1 - 3x)^2}\, dx

=

\displaystyle \int\frac{1}{(2 + x)}\, dx + \displaystyle \int\frac{3}{(1 - 3x)}\, dx + \displaystyle \int\frac{4}{(1 - 3x)^2}\, dx

=

\ln\left|2+x\right| - \ln\left|1 - 3x\right| + \frac{4}{3 - 9x}

For the limits [0, -1]

=

\left[ \ln\left|\frac{2 + x}{1 - 3x}\right| + \frac{4}{3 - 9x} \right]_{-1}^{0}

=

\left[ \ln2 + \frac{4}{3} \right] - \left[\ln\frac{1}{4} + \frac{4}{12}\right]

=

1 + \ln8

Reply 6

wookymon

1

Spot on Dean. For a start the partial fraction solutions are definitely correct. The integration is correct above. I then put the limits of -1 and 0 in, which gives:

(In 2 - In 1) - (In 1 - In 4) + ((4/3) - (4/12))

In 1 = 0 and ((4/3) - (4/12)) = 1

In 2 + In 4 = In 8

Therefore solution is 1 + In 8.

Partial Fractions, what am I doing wrong? (Handwritten pic)

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