1. ## integration problem

hi there,

can anyone help me understand how to work with this problem?

ingration of ( x - 2 ) / sqrt (x - 1)

2. Let u = sqrt(x-1)

then du = 1/(2sqrt(x-1) dx

also u^2 = x-1

so x-2 = u^2 -1

3. $\displaystyle \frac{x-2}{\sqrt{x-1}}=\frac{x-1}{\sqrt{x-1}}-\frac{1}{\sqrt{x-1}}=\sqrt{x-1}-\frac{1}{\sqrt{x-1}}$

The last expression is easy to integrate.

4. thanks spec, that was awesome

5. Hello, weskerq8!

Another approach . . .

$\displaystyle \int \frac{x - 2}{\sqrt{x - 1}}\,dx$

Let: $\displaystyle u \:=\:\sqrt{x-1} \quad\Rightarrow\quad u^2 \:=\:x-1 \quad\Rightarrow\quad x \:=\:u^2+1\quad\Rightarrow\quad dx \:=\:2u\,du$

Substitute: .$\displaystyle \int\frac{u^2-1}{u}(2u\,du) \;=\;2\int(u^2-1)\,du \;=\;2\left(\frac{u^3}{3} - u\right) + C \;=\;\tfrac{2}{3}u(u^2-3) + C$

Back-substitute: . $\displaystyle \tfrac{2}{3}\sqrt{x-1}\bigg[(\sqrt{x-1})^2 - 3\bigg] + C \;=\;\tfrac{2}{3}\sqrt{x-1}\,(x - 1 - 3)$

. . . . . . . . . . . . $\displaystyle = \;\tfrac{2}{3}(x-4)\sqrt{x-1} + C$