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mth 174
Your 'question form' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** Question Form_labelMessages **
partial fractions
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I cannot figure out what I did wrong on this integral using partial fractions the answer in the back of the text is
-15/8 ln (4-x) - ln(x+4)+c but I got
15/8 ln (-x+4) + 15/8 ln (x+4) +c
below are my steps
integral 15/(16-x^2)
15*integral 1/(16-x^2) now factor
15* integral 1/[(-x+4)(x+4)]
remember 15 for later but for now work on partial fractions
A/(-x+4) + B/(x+4) so to get common denominator we multiply
A(x+4) + B(-x+4) gettign Ax + 4A -Bx + 4B =1(numerator)
x(A - B) =0
A-B=0
A=B
4A + 4B = 1 (sub in A=B)
4B + 4B =1
8B=1
B=1/8
going back to integral fill in for A and B
15* integral (1/8) / (-x+4) + 15* integral (1/8)/(x+4)
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My answer agrees with yours up to this point.
However you need to be careful with the substitution you use in integrating 1 / (-x + 4).
When you integrate 1 / (-x+4) you'll let u = -x + 4 so that du = -dx. Your integral will be - ln(|u|) = - ln( |-x+4 | ).
The correct result is then
15 / 8 ( -ln( | -x + 4 | ) ) + ln( |x + 4 | )) + c.
The integrand is not defined at x = 4 or x = -4, at which points the denominator is zero.
Assuming that -4 < x < 4, we have
| x + 4 | = x + 4
and
| -x + 4 | = 4 - x.
In this case our integral is
15 / 8 ( -ln( 4 - x ) ) + ln( x + 4 )) + c
= 15 / 8 * ln ( (x + 4) / (4 - x) ) + c.
Now if x > 4, | -x + 4 | = - (-x + 4) = x - 4 and our integral is
15 / 8 ( -ln( x - 4 ) ) + ln( x + 4 )) + c
= 15 / 8 * ln ( (x + 4) / (x - 4) ) + c.
And if x < -4, our integral is the same:
15 / 8 ( -ln( 4 - x ) ) + ln(-( x + 4) )) + c
= 15 / 8 * ln ( - (x + 4) / (4 - x) ) + c
= 15 / 8 * ln( (x + 4) / (x - 4) )+ c.
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15/8ln (-x+4) + 15/8 (ln (x+4) +c
Can you tell me what I am doing wrong?
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I don't completely agree with you or the book, but I'm closer to your conclusion.
Check my notes.
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