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Integration - hyperbolic functions

If
Unparseable latex formula:

\displaystyle \int \frac{1}{4-x^2} dx = \frac{1}{2} artanh \frac{x}{2} +c =\frac{1}{4} \ln \frac{x+2}{2-x}}



Why doesn't ∫−14−x2dx=−12artanhx2+c=14ln⁡2−xx+2+c \displaystyle \int \frac{-1}{4-x^2} dx =-\frac{1}{2} artanh \frac{x}{2} + c =\frac{1}{4} \ln \frac{2-x}{x+2} +c ?
Reply 1
Avatar for Ano123
Ano123
OP
15
Anyone?????????
Original post by Ano123
Anyone?????????


how do you know that it is wrong ?

:beard:
Reply 3
Avatar for GUMI
GUMI
19
Cuz you can't distribute negative like that (3rd step)
(edited 8 years ago)
Reply 4
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Ano123
OP
15
Original post by GUMI
Cuz you can't distribute negative like that (3rd step)


Why not?
Reply 5
Avatar for Ano123
Ano123
OP
15
Original post by the bear
how do you know that it is wrong ?

:beard:


It doesn't give the same result if you do it by partial fractions (it does work without the negative - it works fine for both methods)
Original post by Ano123
If
Unparseable latex formula:

\displaystyle \int \frac{1}{4-x^2} dx = \frac{1}{2} artanh \frac{x}{2} +c =\frac{1}{4} \ln \frac{x+2}{2-x}}



Why doesn't ∫−14−x2dx=−12artanhx2+c=14ln⁡2−xx+2+c \displaystyle \int \frac{-1}{4-x^2} dx =-\frac{1}{2} artanh \frac{x}{2} + c =\frac{1}{4} \ln \frac{2-x}{x+2} +c ?


I think maybe because ∣2−x∣=∣x−2∣ |2-x| = |x-2| (giving you what is in the formula booklet) so:

∣2−xx+2∣=∣x−2x+2∣ | \dfrac{2-x}{x+2} | = | \dfrac{x-2}{x+2} |

So they both work (I think).
(edited 8 years ago)
Reply 7
Avatar for Zacken
Zacken
22
Original post by Ano123
Anyone?????????


Who say's it's incorrect? That's perfectly fine. Just remember your modulus signs.
Original post by Zacken
Who say's it's incorrect? That's perfectly fine. Just remember your modulus signs.


Yeah I figured that it had to be correct haha.
Reply 9
Avatar for Ayman!
Ayman!
15
It definitely works. You can simply factor out -1.
Reply 10
Avatar for Ano123
Ano123
OP
15
When I do ∫−69−x2dx \int \frac{-6}{9-x^2} dx using hyperbolics i get ln⁡u−3u+3 \ln \frac{u-3}{u+3}
When I do it by partial fractions I get ln⁡3−uu+3 \ln \frac{3-u}{u+3}
Reply 11
Avatar for Ano123
Ano123
OP
15
Is
Unparseable latex formula:

\displaystyle \int \frac{1}{1-x^2} dx =\frac{1}{2} \ln \left \mid \frac{1+x}{1-x} \right \mid

.
Is there an absolute inside the log?
Reply 12
Avatar for Zacken
Zacken
22
Original post by Ano123
When I do ∫−69−x2dx \int \frac{-6}{9-x^2} dx using hyperbolics i get ln⁡u−3u+3 \ln \frac{u-3}{u+3}
When I do it by partial fractions I get ln⁡3−uu+3 \ln \frac{3-u}{u+3}


You don't get either of those, slap on those modulus signs!
Reply 13
Avatar for Zacken
Zacken
22
Original post by Ano123
Is
Unparseable latex formula:

\displaystyle \int \frac{1}{1-x^2} dx =\frac{1}{2} \ln \left \mid \frac{1+x}{1-x} \right \mid

.
Is there an absolute inside the log?


Yes, you can't have the logarithm of a negative number in the real plane.
Reply 14
Avatar for TeeEm
TeeEm
19
Original post by Ano123
When I do ∫−69−x2dx \int \frac{-6}{9-x^2} dx using hyperbolics i get ln⁡u−3u+3 \ln \frac{u-3}{u+3}
When I do it by partial fractions I get ln⁡3−uu+3 \ln \frac{3-u}{u+3}

same answer
you need mods
Reply 15
Avatar for Ano123
Ano123
OP
15
Damn. Makes sense now.
I always used to put the mod signs but for some reason I started thinking that it wasn't right like that.
Original post by TeeEm
same answer
you need mods


How you doing sir
Reply 17
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TeeEm
19
Original post by It'sDuckFam
How you doing sir


very good ..
I am teaching now
Is this further maths
Reply 19
Avatar for B_9710
B_9710
18
Original post by under8ed
Is this further maths


Yes.
Well, hyperbolic functions only appear at further, partial fractions is core 4

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