C
coca cola
Guest
Can you differentiate ln(x + sqrt(x^2 - a^2)), x>a for me without using the standard integrals?
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BlackJack
Vertigo!
In general,
d/dx(ln(f(x))) = f'(x)/f(x) for any function f(x).
Hence,
d/dx(ln(x+sqrt(x^2-a^2)) {note f(x) here is x+sqrt(x^2-a^2).}
= (1+x/sqrt(x^2-a^2))/(x+sqrt(x^2-a^2))
= (sqrt(x^2-a^2)+x)/(sqrt(x^2-a^2)*(x+sqrt(x^2-a^2))
=1/sqrt(x^2-a^2)
...this is what you're looking for?
d/dx(ln(f(x))) = f'(x)/f(x) for any function f(x).
Hence,
d/dx(ln(x+sqrt(x^2-a^2)) {note f(x) here is x+sqrt(x^2-a^2).}
= (1+x/sqrt(x^2-a^2))/(x+sqrt(x^2-a^2))
= (sqrt(x^2-a^2)+x)/(sqrt(x^2-a^2)*(x+sqrt(x^2-a^2))
=1/sqrt(x^2-a^2)
...this is what you're looking for?
C
coca cola
Guest
yep.
Can you integrate it backwards for me: integrate 1/(x^2 - a^2)?
Can you integrate it backwards for me: integrate 1/(x^2 - a^2)?
Grey Council
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- 2004
try doing it by partial fractions
1/(x-a)(x+a)
1/(x-a)(x+a)
thats meant to be in square root isnt it?coca cola said:
alternative to no_arg's:
let x = a*sec@
then, after simplifying, you'll have to do int sec@ d@
which can be done by writing sec@ as
sec@ (sec@ + tan@) / (sec@ + tan @)
= (sec^2@ + sec@tan@) / (tan@ + sec@)
not that its easier
but if you dont have the standard integral sheet at all and you somehow forget that it will have "x + sqrt(x^2-a^2)", this method may be a longer alternative.