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

[whatlifeforme]

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.

 

[whatlifeforme]

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

 

[Dick]

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.