integrate 1/((x)^(1/2)+9))

books answer is 2x^(1/2)-log((x)^(1/2)+9) + c

my question is can you also integrate this as 1/((x)^(1/4)^(2))+3^(2) to get (1/3) inverse tan x/3 + c????

are these different versions of the same answer???

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- Sep 22nd 2012, 07:51 AMsluggerbrothlog vs inverse tan?????
integrate 1/((x)^(1/2)+9))

books answer is 2x^(1/2)-log((x)^(1/2)+9) + c

my question is can you also integrate this as 1/((x)^(1/4)^(2))+3^(2) to get (1/3) inverse tan x/3 + c????

are these different versions of the same answer??? - Sep 22nd 2012, 08:35 AMFernandoRevillaRe: log vs inverse tan?????
- Sep 22nd 2012, 09:39 AMsluggerbrothRe: log vs inverse tan?????
are you saying (1/3)inverse tan x^(1/4)/3 is the same as books answer

of 2x^(1/2)-log((x)^(1/2)+9) + c - Sep 22nd 2012, 09:52 AMSworDRe: log vs inverse tan?????
No you cannot use inverse tan here... the power at the bottom is 1/2, NOT 2.

You have

Make the substitution . This will lead you to a quotient of two linear equations; simplify them using polynomial division, and then the result should be easy to integrate. - Sep 22nd 2012, 10:07 AMsluggerbrothRe: log vs inverse tan?????
If you re wrote this as 1/ (((x)^(1/4))^(2)+3^(2))

i see how it works your way but why would the way have it above NOT WORK? It looks just like the integral of 1/(a^(2)+U^(2))=1/a (inverse tan u/a) +c

what am i missing??/ - Sep 22nd 2012, 11:13 AMSworDRe: log vs inverse tan?????
No because you need the derivative of the inside to apply the chain rule.

For example if

then

But in our case the derivative of the inside isn't available so inverse tan won't work..