can i start by rewriting it as

= (x^-1/2)(1 +√x)^-2??

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- Dec 14th 2012, 09:22 AMasilvester635Integration help
can i start by rewriting it as

= (x^-1/2)(1 +√x)^-2?? - Dec 14th 2012, 09:41 AMskeeterRe: Integration help
substitution ...

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

$\displaystyle du = \frac{1}{2\sqrt{x}} \, dx$ - Dec 14th 2012, 09:44 AMasilvester635Re: Integration help
so when i derive 1 + √x does the square root turn into a negative ^-1/2 or will it be positive??

- Dec 14th 2012, 10:15 AMskeeterRe: Integration help
- Dec 14th 2012, 10:50 AMPlatoRe: Integration help
- Dec 17th 2012, 05:46 PMx3bnmRe: Integration help
$\displaystyle \text{Let } u = 1 + \sqrt{x} \text{ then } \frac{du}{dx} = \frac{1}{2\sqrt{x}}$

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