1. ## Integration help

can i start by rewriting it as

= (x^-1/2)(1 +√x)^-2??

2. ## Re: Integration help

substitution ...

$\displaystyle u = 1 + \sqrt{x}$

$\displaystyle du = \frac{1}{2\sqrt{x}} \, dx$

3. ## Re: Integration help

so when i derive 1 + √x does the square root turn into a negative ^-1/2 or will it be positive??

4. ## Re: Integration help

Originally Posted by asilvester635
so when i derive 1 + √x does the square root turn into a negative ^-1/2 or will it be positive??
look at my response again ...

btw, it's differentiate, not "derive".

5. ## Re: Integration help

Originally Posted by asilvester635
so when i derive 1 + √x does the square root turn into a negative ^-1/2 or will it be positive??

What is the derivative of $\displaystyle -2\left(1+\sqrt{x}\right)^{-1}~?$

6. ## Re: Integration help

Originally Posted by asilvester635
can i start by rewriting it as

= (x^-1/2)(1 +√x)^-2??
$\displaystyle \text{Let } u = 1 + \sqrt{x} \text{ then } \frac{du}{dx} = \frac{1}{2\sqrt{x}}$

And so $\displaystyle 2du = \frac{1}{\sqrt{x}} dx$

By plugging in $\displaystyle u = 1 + \sqrt{x}$ and $\frac{1}{\sqrt{x}} dx = 2 du$ into $\displaystyle \int_1^9 \frac{1}{\sqrt{x}(1+\sqrt{x})^2} dx$ we get:

\displaystyle \begin{align*}\int_1^9 \frac{1}{\sqrt{x}(1+\sqrt{x})^2} dx =& \int_1^9 \frac{2}{u^2} du\\ =& \frac{-2}{u}\Big{]}_1^9 \\ =& \frac{-2}{1+\sqrt{x}}\Big{]}_1^9 \\ =& \frac{-2}{1+3} + \frac{2}{1+\sqrt{1}} \\ =& \frac{-2}{4} + \frac{2}{2} \\ =& \frac{-1}{2}+1 \\ =& \frac{1}{2}\end{align*}

$\displaystyle \therefore \int_1^9 \frac{1}{\sqrt{x}(1+\sqrt{x})^2} dx = \frac{1}{2}$

Check:

integration 1/(((x)^(1/2))((1+((x)^(1/2))))^2) dx from x=1 to 9 - Wolfram|Alpha