# Solutions to Problem Set 7

Math 211-03

9-18-2017

[Miscellaneous substitutions]

1. Compute .

The quadratic does not factor. The middle term in the quadratic is . Since and , I have

Therefore,

2. Compute .

Since factors into , I'll use partial fractions. (You could do this by completing the square, then doing a trig substitution. I think it would be a little longer.)

Write

Let . I get , so .

Let . I get , so .

Therefore,

3. Compute .

While , factoring doesn't help. I'll complete the square. The middle term is . Since and , I have

Therefore,

4. Compute .

The quadratic does not factor. I'll use partial fractions.

Let . I get , so . Then

Let . I get , so . Then

At this point, I can reduce the equation by differentiating both sides, or I can plug another value in for x. I'll let . I get

Therefore,

For the first term, I have

For the second term, I complete the square:

(The last step follows by making the easy substitution . You can work out the details.)

Therefore, the answer to the original question is

5. Compute .

6. Compute .

Since I see lots of powers of , I'll use :

Apply partial fractions:

Let . I get , so .

Let . I get , so .

Therefore,

7. Compute .

Here's the work for the long division:

Those who reach greatness on earth reach it through concentration. - The Upanishads

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