## ${\int}_{0}^{2} \frac{x}{\sqrt{1 + {x}^{2}}} \mathrm{dx}$

Nov 26, 2017

${\int}_{0}^{2} \frac{x}{\sqrt{1 + {x}^{2}}} \cdot \mathrm{dx} = \sqrt{5} - 1$

#### Explanation:

${\int}_{0}^{2} \frac{x}{\sqrt{1 + {x}^{2}}} \cdot \mathrm{dx}$

=${\int}_{0}^{2} \frac{2 x \cdot \mathrm{dx}}{2 \sqrt{1 + {x}^{2}}}$

After using $u = {x}^{2}$ and $\mathrm{du} = 2 x \cdot \mathrm{dx}$ transforms, this integral became

${\int}_{0}^{4} \frac{\mathrm{du}}{2 \sqrt{1 + u}}$

=$\frac{1}{2} \cdot {\int}_{0}^{4} {\left(1 + u\right)}^{- \frac{1}{2}} \cdot \mathrm{du}$

=$\frac{1}{2} \cdot {\left[{\left(1 + u\right)}^{- \frac{1}{2} + 1} / \left(- \frac{1}{2} + 1\right)\right]}_{0}^{4}$

=$\frac{1}{2} \cdot {\left[{\left(1 + u\right)}^{\frac{1}{2}} / \left(\frac{1}{2}\right)\right]}_{0}^{4}$

=${\left[{\left(1 + u\right)}^{\frac{1}{2}}\right]}_{0}^{4}$

=${5}^{\frac{1}{2}} - {1}^{\frac{1}{2}}$

=$\sqrt{5} - 1$

Nov 26, 2017

${\int}_{0}^{2} \frac{x}{\sqrt{1 + {x}^{2}}} \mathrm{dx} = \sqrt{5} - 1$

#### Explanation:

let $u = {x}^{2} + 1$, then $\mathrm{du} = 2 x \mathrm{dx}$ or $x \mathrm{dx} = \frac{1}{2} \mathrm{du}$

Change the limits:

$a = {0}^{2} + 1$

$a = 1$

$b = {2}^{2} + 1$

$b = 5$

${\int}_{0}^{2} \frac{x}{\sqrt{1 + {x}^{2}}} \mathrm{dx} = \frac{1}{2} {\int}_{1}^{5} {u}^{- \frac{1}{2}} \mathrm{du}$

${\int}_{0}^{2} \frac{x}{\sqrt{1 + {x}^{2}}} \mathrm{dx} = {\left({u}^{\frac{1}{2}}\right]}_{1}^{5}$

${\int}_{0}^{2} \frac{x}{\sqrt{1 + {x}^{2}}} \mathrm{dx} = \sqrt{5} - 1$

Nov 26, 2017

${\int}_{0}^{2} \frac{x}{\sqrt{1 + {x}^{2}}} \mathrm{dx} = \sqrt{5} - 1$

#### Explanation:

We can make a substitution to solve this integral:

Let $u = 1 + {x}^{2}$

${\int}_{0}^{2} \frac{x}{\sqrt{u}} \mathrm{dx}$

And so $\mathrm{du} = 2 x \mathrm{dx}$ which we can then solve for $x$ to get $\frac{1}{2} \mathrm{du} = x \mathrm{dx}$

Replacing $x \mathrm{dx}$ with $\frac{1}{2} \mathrm{du}$ we now have

${\int}_{0}^{2} \frac{1}{2 \sqrt{u}} \mathrm{du}$

Take out the constant...

$\frac{1}{2} {\int}_{0}^{2} \frac{1}{\sqrt{u}} \mathrm{du}$

Applying the power rule for integration:

$\frac{1}{2} {\int}_{0}^{2} {\left(u\right)}^{- \frac{1}{2}} \mathrm{du}$

$\frac{1}{2} \cdot \frac{{\left(u\right)}^{\left(- \frac{1}{2}\right) + 1}}{\left(- \frac{1}{2}\right) + 1}$

$\frac{1}{2} \cdot {\left(u\right)}^{\frac{1}{2}} / \left(\frac{1}{2}\right)$

$\frac{1}{2} \cdot 2 \sqrt{u}$

Simplify and substitute back $u = 1 + {x}^{2}$

${\left[\sqrt{1 + {x}^{2}} + \text{C}\right]}_{0}^{2}$

Now evaluating the integral from $0$ to $2$

$\left[\sqrt{1 + {\textcolor{red}{2}}^{2}} + \text{C"] - [sqrt(1+color(red)0^2)+"C}\right]$

$\left[\sqrt{5}\right] - \left[1\right]$

So

${\int}_{0}^{2} \frac{x}{\sqrt{1 + {x}^{2}}} \mathrm{dx} = \sqrt{5} - 1$