# Thread: Find the Definite Integral

1. ## Find the Definite Integral

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.

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

3. Maybe have a closer look at your quesiton as,

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

4. 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)

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

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

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

Therefore

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

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

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

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

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

Now long divide to get

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

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

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

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

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

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

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