log 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???

Quote:

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Originally Posted by sluggerbroth

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

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We have:



But we need in the numerator the derivative of

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

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.

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??/

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..