and

so that

and .

Therefore,

.

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* SOLUTION 2 :* Integrate
. Let

and

so that

and .

Therefore,

.

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* SOLUTION 3 :* Integrate
. Let

and

so that

and .

Therefore,

.

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* SOLUTION 4 :* Integrate
. Let

and

so that

and .

Therefore,

.

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* SOLUTION 5 :* Integrate
. Let

and

so that

and .

Therefore,

.

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* SOLUTION 6 :* Integrate
. Let

and

so that (Don't forget to use the chain rule when differentiating .)

and .

Therefore,

.

Now use u-substitution. Let

so that

,

or

.

Then

+ C

+ C

+ C .

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* SOLUTION 7 :* Integrate
. Let

and

so that

and .

Therefore,

.

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* SOLUTION 8 :* Integrate
. Let

and

so that

and .

Therefore,

(Add in the numerator. This will replicate the denominator and allow us to split the function into two parts.)

.

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* SOLUTION 9 :* Integrate
. Let

and

so that

and .

Therefore,

.

Integrate by parts again. Let

and

so that

and .

Hence,

.

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Duane Kouba 2000-04-23