# log vs inverse tan?????

• September 22nd 2012, 07:51 AM
sluggerbroth
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???
• September 22nd 2012, 08:35 AM
FernandoRevilla
Re: log vs inverse tan?????
Quote:

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????

We have:

$\displaystyle\int \frac{dx}{x^{1/2}+9}=\displaystyle\int \frac{dx}{(x^{1/4})^2+3^2}=\frac{1}{9}\displaystyle\int \frac{dx}{\left(\frac{x^{1/4}}{3}\right)^2+1}$

But we need in the numerator the derivative of $\frac{x^{1/4}}{3}.$
• September 22nd 2012, 09:39 AM
sluggerbroth
Re: 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
• September 22nd 2012, 09:52 AM
SworD
Re: log vs inverse tan?????
No you cannot use inverse tan here... the power at the bottom is 1/2, NOT 2.

You have

$\frac{1}{\sqrt{x}+9}$

Make the substitution $x = u^2$. 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.
• September 22nd 2012, 10:07 AM
sluggerbroth
Re: 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??/
• September 22nd 2012, 11:13 AM
SworD
Re: log vs inverse tan?????
No because you need the derivative of the inside to apply the chain rule.

For example if

$f(x) = \tan^{-1}({3x^2})$ then

$f'(x) = \frac{6x}{1+(3x^2)^2}$

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