Integrate x^3/(x^2 - 16) with Trig Substitution

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SUMMARY

The integral of x^3/(x^2 - 16) can be evaluated using trigonometric substitution, specifically with the substitution x = 4sec(θ). The correct answer is (x^2)/2 + 8 * ln|x^2 - 16| + C, while a common mistake involves an additional -8 in the solution. Both solutions yield the same derivative, confirming their correctness, as the constant can be absorbed into the arbitrary constant C.

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whatlifeforme
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Homework Statement


evaluate the integral.


Homework Equations


integral (x^3 / (x^2 - 16)


The Attempt at a Solution


x=4sec∅
dx=4sec∅tan∅d∅

1. i substituted those values in, and then split sec^4∅ into sec^2∅ and (1+tan^2∅).
2. integral 16 (1/u) du + integral 16 (u) du.
3. end with:

16 * ln|sqrt(x^2 - 16) / 4| + (1/2) (x^2 - 16)


however, the answer is:

(x^2)/2 + 8 * ln|x^2 - 16| + c.
 
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okay. i tried u=x^2 -16

1/2 integral (u+16) / u du

1/2 integral (du) + 1/2 integral (16/u) du

1/2(x^2 - 16) + 8 * ln|x^2 - 16|

i get part of the correct answer: (x^2 / 2) + 8 * ln|x^2 - 16| but i have an extra -8 .

correct answer: (x^2 / 2) + 8 * ln|x^2 - 16| + c
my answer: (x^2 / 2) + 8 * ln|x^2 - 16| - 8 + c
 
whatlifeforme said:
okay. i tried u=x^2 -16

1/2 integral (u+16) / u du

1/2 integral (du) + 1/2 integral (16/u) du

1/2(x^2 - 16) + 8 * ln|x^2 - 16|

i get part of the correct answer: (x^2 / 2) + 8 * ln|x^2 - 16| but i have an extra -8 .

correct answer: (x^2 / 2) + 8 * ln|x^2 - 16| + c
my answer: (x^2 / 2) + 8 * ln|x^2 - 16| - 8 + c

If you differentiate both of those you get the same thing. They are both correct. You can absorb the -8 into the +c.
 

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