# How do we determine the value of int x/sqrt(x - 2) dx?

Feb 2, 2017

The integral equals $2 x {\left(x - 2\right)}^{\frac{1}{2}} - \frac{4}{3} {\left(x - 2\right)}^{\frac{3}{2}} + C$

#### Explanation:

First rewrite the integral as a product.

$\int \frac{x}{\sqrt{x - 2}} \mathrm{dx} = \int \frac{x}{x - 2} ^ \left(\frac{1}{2}\right) \mathrm{dx} = \int x {\left(x - 2\right)}^{- \frac{1}{2}} \mathrm{dx}$

Now let $u = x$ and $\mathrm{dv} = {\left(x - 2\right)}^{- \frac{1}{2}} \mathrm{dx}$. We instantly know that $\mathrm{du} = \mathrm{dx}$. To integrate $\mathrm{dv}$, we need to make a substitution. Let $n = x - 2$. Then $\mathrm{dn} = \mathrm{dx}$.

$\int {\left(x - 2\right)}^{- \frac{1}{2}} \mathrm{dx} = \int {n}^{- \frac{1}{2}} \mathrm{dn} = 2 {n}^{\frac{1}{2}} = 2 {\left(x - 2\right)}^{\frac{1}{2}}$

So, to summarize:

$\left\{\begin{matrix}u = x \\ \mathrm{du} = \mathrm{dx} \\ \mathrm{dv} = {\left(x - 2\right)}^{- \frac{1}{2}} \mathrm{dx} \\ v = 2 {\left(x - 2\right)}^{\frac{1}{2}}\end{matrix}\right.$

We can add the constant of integration at the end.

Now use the above values in the integration by parts formula, which is:

$\int u \mathrm{dv} = u v - \int v \mathrm{du}$

$\int x {\left(x - 2\right)}^{- \frac{1}{2}} \mathrm{dx} = x \left(2 {\left(x - 2\right)}^{\frac{1}{2}}\right) - \int 2 {\left(x - 2\right)}^{\frac{1}{2}} \mathrm{dx}$

$\int x {\left(x - 2\right)}^{- \frac{1}{2}} \mathrm{dx} = 2 x {\left(x - 2\right)}^{\frac{1}{2}} - 2 \int {\left(x - 2\right)}^{\frac{1}{2}}$

We solve this using another substitution. Let $m = x - 2$. Then $\mathrm{dm} = \mathrm{dx}$.

$\int {m}^{\frac{1}{2}} \mathrm{dm} = \frac{2}{3} {m}^{\frac{3}{2}} = \frac{2}{3} {\left(x - 2\right)}^{\frac{3}{2}}$

Therefore, we have:

$\int x {\left(x - 2\right)}^{- \frac{1}{2}} \mathrm{dx} = 2 x {\left(x - 2\right)}^{\frac{1}{2}} - \frac{4}{3} {\left(x - 2\right)}^{\frac{3}{2}} + C$

Hopefully this helps!

Feb 2, 2017

$= \frac{2}{3} \sqrt{x - 2} \left(x + 4\right) + C$

#### Explanation:

$\int \frac{x}{\sqrt{x - 2}} \mathrm{dx}$

Using a substitution, we could say that: $u = \sqrt{x - 2}$ so that $x = {u}^{2} + 2 = \implies \mathrm{dx} = 2 u \mathrm{du}$

The integration then becomes:

$\int \frac{{u}^{2} + 2}{u} \cdot 2 u \mathrm{du}$

$= 2 \int {u}^{2} + 2 \mathrm{du} = \frac{2 {u}^{3}}{3} + 4 u + C$

If we undo the sub, we have:

$= \frac{2 {\left(\sqrt{x - 2}\right)}^{3}}{3} + 4 \sqrt{x - 2} + C$

$= \sqrt{x - 2} \left(\frac{2 x - 4}{3} + 4\right) + C$

$= \sqrt{x - 2} \left(\frac{2 x}{3} + \frac{8}{3}\right) + C$

$= \frac{2}{3} \sqrt{x - 2} \left(x + 4\right) + C$

With IBP we can recognise that:

$\int \frac{x}{\sqrt{x - 2}} \mathrm{dx} = \int x \cdot \frac{1}{\sqrt{x - 2}} \mathrm{dx} = \int x \cdot {\left(2 \sqrt{x - 2}\right)}^{p} r i m e \mathrm{dx}$

[this is because ${\left(2 \sqrt{x - 2}\right)}^{p} r i m e = \frac{1}{\sqrt{x - 2}}$]

$= 2 x \sqrt{x - 2} - \int {\left(x\right)}^{p} r i m e 2 \sqrt{x - 2} \mathrm{dx}$

$= 2 x \sqrt{x - 2} - 2 \int \sqrt{x - 2} \mathrm{dx}$

$= 2 x \sqrt{x - 2} - 2 \cdot \frac{2}{3} {\left(x - 2\right)}^{\frac{3}{2}} + C$

$= \sqrt{x - 2} \left(2 x - 2 \cdot \frac{2}{3} \left(x - 2\right)\right) + C$

$= \sqrt{x - 2} \left(\frac{2 x}{3} + \frac{8}{3}\right) + C$

$= \frac{2}{3} \sqrt{x - 2} \left(x + 4\right) + C$

Same thing!