query 17

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course Mth 272

7/8 1:31pm

017.

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Question: `qQuery problem 6.2.21 (7th edition 6.3.18) integrate 3/(x ^ 2 - 3x)

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Your solution:

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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Given Solution:

`a 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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Question: `qQuery problem 6.2.27 (7th edition 6.3.29) (was 6.3.27) integrate (x+2) / (x^2 - 4x)

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Your solution:

We factor x out of the denominator to get

(x+2)/ [ x(x-4) ]

Use partial fractions:

(x+2)/x(x-4) = A/x + B/(x-4)

Multiply both sides by common denominator:

x+2 = A(x-4) + B(x) or

x+2 = (A+B) x - 4 A.

Thus

A + B = 1 and

-4 A = 2 so

A = -1/2 and

B = 3/2.

Our integrand becomes

(-1/2) / x + (3/2) / (x-4). The general antiderivative is easily found to be

3/2 ln |x-4| - 1/2 ln |x| + c, which can be expressed as

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

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

`a We factor x out of the denominator to get

(x+2)/ [ x(x-4) ]

Use partial fractions:

(x+2)/x(x-4) = A/x + B/(x-4)

Multiply both sides by common denominator:

x+2 = A(x-4) + B(x) or

x+2 = (A+B) x - 4 A.

Thus

A + B = 1 and

-4 A = 2 so

A = -1/2 and

B = 3/2.

Our integrand becomes

(-1/2) / x + (3/2) / (x-4). The general antiderivative is easily found to be

3/2 ln |x-4| - 1/2 ln |x| + c, which can be expressed as

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

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Question: `qQuery Add comments on any surprises or insights you experienced as a result of this assignment."

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Question: `qQuery Add comments on any surprises or insights you experienced as a result of this assignment."

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&#Good responses. Let me know if you have questions. &#