Integration help

**asilvester635** · December 14th 2012, 09:22 AM

can i start by rewriting it as

= (x^-1/2)(1 +√x)^-2??

**skeeter** · December 14th 2012, 09:41 AM

substitution ...

$u = 1 + \sqrt{x}$

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

**asilvester635** · December 14th 2012, 09:44 AM

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

**skeeter** · December 14th 2012, 10:15 AM

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".

**Plato** · December 14th 2012, 10:50 AM

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 $-2\left(1+\sqrt{x}\right)^{-1}~?$

**x3bnm** · December 17th 2012, 05:46 PM

Originally Posted by asilvester635

can i start by rewriting it as

= (x^-1/2)(1 +√x)^-2??

$\text{Let } u = 1 + \sqrt{x} \text{ then } \frac{du}{dx} = \frac{1}{2\sqrt{x}}$

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

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

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

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

Check:

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

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