Find the Definite Integral

**ugkwan** · September 24th 2010, 06:04 PM

Hi

(intetral sign) 1/ (1+ 2^(1/2)) dx

I subsituted u= 1 + 2^(1/2) but got no where after du = x^(-1/2) dx

Algebraic help or hints please.

**Prove It** · September 24th 2010, 06:22 PM

$\displaystyle{\frac{1}{1 + 2^{\frac{1}{2}}}}$ is a constant. What does that tell you about its antiderivative?

**pickslides** · September 24th 2010, 06:23 PM

Maybe have a closer look at your quesiton as,

$\displaystyle \int \frac{1}{1+2^{\frac{1}{2}}}~dx =\frac{x}{1+2^{\frac{1}{2}}}+C$

**ugkwan** · September 24th 2010, 06:56 PM

I am so sorry that I incorrectly posted the problem. I should of taken the time to double check instead of wasting your time.

The question is ∫ 1/ (1 + √(2x) )

Again I substituted 1 + √(2x) and got stuck with du = 1/√(x)

**Prove It** · September 24th 2010, 07:17 PM

$\frac{1}{1 + \sqrt{2x}} = \frac{\sqrt{x}}{\sqrt{x}(1 + \sqrt{2x})}$

$= \frac{\sqrt{x}}{1 + \sqrt{2}\sqrt{x}}\left(\frac{1}{\sqrt{x}}\right)$

$= \frac{2\sqrt{x}}{1 + \sqrt{2}\sqrt{x}}\left(\frac{1}{2\sqrt{x}}\right)$ .

Therefore

$\int{\frac{1}{1 + \sqrt{2x}}\,dx} = \int{\frac{2\sqrt{x}}{1 + \sqrt{2}\sqrt{x}}\left(\frac{1}{2\sqrt{x}}\right)\ ,dx}$ .

Now let $u = \sqrt{x}$ so that $\frac{du}{dx} = \frac{1}{2\sqrt{x}}$ so that

$\int{\frac{2\sqrt{x}}{1 + \sqrt{2}\sqrt{x}}\left(\frac{1}{2\sqrt{x}}\right)\ ,dx} = \int{\frac{2u}{1 + \sqrt{2}u}\,\frac{du}{dx}\,dx}$

$= \int{\frac{2u}{1 + \sqrt{2}u}\,du}$

$= 2\int{\frac{u}{1 + \sqrt{2}u}\,du}$ .

Now long divide to get

$= 2\int{\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}(1 + \sqrt{2}u)}\,du}$

$= 2\left[\frac{u}{\sqrt{2}} - \frac{1}{2}\ln{(1 + \sqrt{2}u)}\right] + C$

$= \sqrt{2}u - \ln{(1 + \sqrt{2}u)} + C$

$= \sqrt{2}\sqrt{x} - \ln{(1 + \sqrt{2}\sqrt{x})} + C$

$= \sqrt{2x} - \ln{(1 + \sqrt{2x})} + C$ .

**TheCoffeeMachine** · September 24th 2010, 07:41 PM

Shorter way: let $\displaystyle u = 1+\sqrt{2x} = 1+\sqrt{2}\sqrt{x}$ , then $\frac{du}{dx} = \frac{\sqrt{2}}{2\sqrt{x}} \Rightarrow {dx} = \frac{2\sqrt{x}}{\sqrt{2}}\;{du}$ .
We have then $\displaystyle \frac{2}{\sqrt{2}}\int \frac{\sqrt{x}}{u}\;{du}$ . But from $u = 1+\sqrt{2x} = 1+\sqrt{2}\sqrt{x}$ , we get $\sqrt{x} = \frac{u-1}{\sqrt{2}}$ .
So we have $\displaystyle \frac{2}{\sqrt{2}\sqrt{2}}\int\frac{u-1}{u}\;{du} = \int\left(1-\frac{1}{u}\right)\;{du} = u-\ln{u}+k.$

As $\displaystyle u = 1+\sqrt{2x}$ , this is $\displaystyle 1+\sqrt{2x}-\ln(1+\sqrt{2x})+k$ .

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