# livinginmyownreality's question at Yahoo! Answers regarding an indefinite integral

#### MarkFL

Staff member
Here is the question:

What is the integral of sqrt(x-1)/x. Show Work.?

Answer should be 2sqrt(x-1) - 2arctan*sqrt(x-1) + C

Use substitution.
I have posted a link there to this topic so the OP can see my work.

#### MarkFL

Staff member
Hello livinginmyownreality,

We are given to evaluate:

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

Using the substitution:

$$\displaystyle u=\sqrt{x-1}\,\therefore\,du=\frac{1}{2\sqrt{x-1}}\,dx\,therefore\,dx=2u\,du$$

$$\displaystyle u^2=x-1\implies x=u^2+1$$

And so we obtain:

$$\displaystyle I=2\int\frac{u^2}{u^2+1}\,du=2\int\frac{u^2+1-1}{u^2+1}\,du=2\int 1-\frac{1}{u^2+1}\,du$$

From this we obtain:

$$\displaystyle I=2\left(u-\tan^{-1}(u) \right)+C$$

Back-substituting for $u$, and distributing the $2$, we get:

$$\displaystyle I=2\sqrt{x-1}-2\tan^{-1}\left(\sqrt{x-1} \right)+C$$