asst 17

Asst # 17

07-12-2006

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11:37:09

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11:37:10

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**** Query problem 6.3.18 integrate 3/(x ^ 2 - 3x)

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12:14:04

ln 3x-9/x + C

Be careful to group your numerators and denominators.

Solution for comparison:

&& First we factor x the demoninator:

3/ [ x(x-3) ]

Use partial fractions:

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

Multiply both sides by common denominator x(x-3) to get

3 = A(x-3) + B(x) or
3 = (A+B) x - 3 A, which is the same as

0 x + 3 = (A + B) x - 3 A.

The coefficients of x on both sides must be the same so we have

A + B = 0 (coefficients of x) and
-3 A = 3 .

From the second we get A = -1. Substituting this into the first we solve to get B = 1.

So our integrand is

3 / (x^2 - 3x) = -1 / (x-3) + 1 / x.

The integration is straightforward. We get

ln |x-3| - ln |x| + c , which we rewrite using the laws of logarithms as

ln | (x-3) / x | + c. **

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12:14:05

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**** What is your integral?

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12:17:08

3/ x(x-3) dx ---> ln 3x-9 / x+ c

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12:17:08

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**** Explain how you used partial fractions to obtain your result.

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12:23:06

A/ ax + b

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12:23:07

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**** Query problem 6.3.27 (was 6.3.29) integrate (x+2) / (x^2 - 4x)

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12:24:26

1/2(3ln x-4 - lnx) + c

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12:24:27

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**** What is your integral?

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12:27:49

x+2/ x(x-4) ---> 1/293lnx-4 - lnx) + C

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12:27:50

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**** Explain how you used partial fractions to obtain your result.

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12:28:07

a/ ax+b - b/ ax+c

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12:28:08