- #1

- 81

- 1

## Homework Statement

[tex]\int \sqrt{a-\frac{1}{x}}dx[/tex]

## 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

$$\sqrt{ax^2-x}+\frac{3}{\sqrt{a}} \ln |\sqrt{ax}+\sqrt{ax-1}|+C$$