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???
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???
No you cannot use inverse tan here... the power at the bottom is 1/2, NOT 2.
You have
$\displaystyle \frac{1}{\sqrt{x}+9}$
Make the substitution $\displaystyle x = u^2$. 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.
No because you need the derivative of the inside to apply the chain rule.
For example if
$\displaystyle f(x) = \tan^{-1}({3x^2})$ then
$\displaystyle f'(x) = \frac{6x}{1+(3x^2)^2}$
But in our case the derivative of the inside isn't available so inverse tan won't work..