Integrating a Radical Function: Is There More Than One Way?

[tempneff]

Homework Statement

$$\int \sqrt{a-\frac{1}{x}}dx$$

Homework Equations

U-sub
Trig Sub

The Attempt at a Solution

I checked the integral with values: a=2, from x=1 to x=2...but it gave a different value then I get with mine. They did it a far different way to, but I can't see why this won't work...can you?


Evaluate:$$\int \sqrt{a-\frac{1}{x}}dx$$
$$\int \sqrt{a-\frac{1}{x}}dx \rightarrow \int \sqrt{\frac{ax-1}{x}}dx\rightarrow \int \frac{\sqrt{ax-1}}{\sqrt{x}}\rightarrow \int \sqrt{\frac{1}{x}}\sqrt{ax-1}$$
Let $$u=\sqrt{ax-1} \hspace{15pt}u^2=ax-1\hspace{15pt}x=\frac{u^2+1}{a}\hspace{15pt}\sqrt{ \frac{1}{x}}=\sqrt { \frac{a}{u^2+1}}$$
and $$du=\frac{a}{2\sqrt{ax-1}} dx\rightarrow dx=\frac{2\sqrt{ax-1}}{a} du \rightarrow \frac{2u}{a}du$$
Now
$$\int \sqrt{a-\frac{1}{x}}dx\rightarrow \int \frac{2\sqrt{a}}{a} \frac{u^2}{\sqrt{u^2+1}} du\rightarrow \frac{2}{\sqrt{a}}\int \frac{u^2}{\sqrt{u^2+1}}du$$
Let $$ u=\tan \theta \hspace{15pt}\sqrt{u^2+1}= \sec \theta $$
and
$$ du =\sec^2 \theta d \theta$$
so
$$ \frac{2}{\sqrt{a}}\int \frac{u^2}{\sqrt{u^2+1}}du \rightarrow \frac{2}{\sqrt{a}}\int \frac{\tan^2 \theta}{\sqrt{\tan^2 \theta +1}}\sec^2 \theta d \theta \rightarrow \frac{2}{\sqrt{a}}\int \frac{\tan^2 \theta}{\sqrt{\tan^2 \theta +1}}\sec^2 \theta d \theta$$
$$ \frac{2}{\sqrt{a}}\int \frac{\sec^2 \theta +1}{\sqrt{\sec^2 \theta}}\sec^2 \theta \rightarrow \frac{2}{\sqrt{a}}\int \sec^3 \theta + \sec \theta d \theta$$
By Integral Table $\# 82$ and $ \#84$
$$=\frac{2}{\sqrt{a}}\left( \frac{1}{2}\sec \theta \tan \theta +\frac{3
}{2} \ln |\sec \theta + \tan \theta |+C \right) $$
where $$ u=\tan \theta \hspace{15pt}\sqrt{u^2+1}= \sec \theta $$
so $$\frac{2}{\sqrt{a}} \left( \frac{1}{2}u\sqrt{u^2+1}+\frac{3}{2}\ln | \sqrt{u^2+1}+u|+C \right)$$
where
$$u=\sqrt{ax-1}$$
so
$$\frac{2}{\sqrt{a}}\left( \frac{1}{2}\sqrt{ax-1}\sqrt{ax}+\frac{3}{2} \ln |\sqrt{ax}+\sqrt{ax-1}|+C\right)$$
or
$$\sqrt{ax^2-x}+\frac{3}{\sqrt{a}} \ln |\sqrt{ax}+\sqrt{ax-1}|+C$$

 

[Jorriss]

I take back my comment. It may not work. You could try u = a - 1/x but I don't know if it works out fully. This isn't integral I thought it was at first.

 

[tempneff]

Yeah I saw it completed with $$u=\frac{ax-1}{x}$$ and I will go that way if I have to, but I really want to know why this isn't working. I broke a rule somewhere...

 

[tt2348]

maybe try x=(sec(theta)^2)/a, that might clean up some of it

 

[andrien]

check out,you have put tan^2(theta)=sec^2(theta)+1, it will be sec^2(theta)-1.just above where you have written 'by integral table'.

 

[tempneff]

Excellent that did it!