Find $\int\sqrt{(x-2)/(x-1)}\,dx$

Question

I am having difficulty with these type of problems.

Can anyone explain how to approach the problems of the form $\int \sqrt{\dfrac{x-a}{x-b}}\hspace{1mm}dx$

Answer 1

You can check by differentiation that \begin{align*} \int\sqrt{\frac{x-a}{x-b}}\,\mathrm dx=\sqrt{(x-a)(x-b)}+(b-a)\ln\left(\sqrt{x-a}+\sqrt{x-b}\right). \end{align*} Of course, one has to careful to define the integrand for only such $x\in\mathbb R$ that $x\neq b$ and $(x-a)/(x-b)\geq0$

Answer 2

For the integral in your title you can substitute $$x-1=\cosh^2\theta\ ,\quad x-2=\cosh^2\theta-1=\sinh^2\theta\ ,\quad dx=2\cosh\theta\sinh\theta\,d\theta\ .$$ Then you get $$I=\int\frac{\sinh\theta}{\cosh\theta}2\cosh\theta\sinh\theta\,d\theta =\int2\sinh^2\theta\,d\theta=\int(\cosh2\theta-1)\,d\theta$$ which is now a standard integral. The generalisation can be done in a similar way.

Answer 3

You could rewrite the integral as

$$\int\sqrt{1-\frac1{x-1}}dx$$

Similar to other trigonometric substitutions, one could eliminate the square root with a trigonometric substitution

$$\frac1{x-1}=\sin^2t,x-1=\csc^2t,dx=-2\csc^2t\cot tdt$$ $$\int\sqrt{1-\frac1{x-1}}dx=\int-2\csc^2t\cot t\sqrt{1-\sin^2t}dt=\int-2\csc t\cot^2tdt=$$ $$\int-2\csc^3tdt+\int2\csc tdt$$

I'm sure you've handled similar integrals with secant.

Answer 4

Let $x-1=\sec^2 u$ so that $dx=2\sec^2u \tan u \ \ \text{and } x-2=\sec^2 u-1$ this shows that $$\int\sqrt\frac{x-2}{x-1}\,dx=2\int \tan^2 u\sec u\, du=2\int (sec^3 u- \sec u)\, du.$$ know, for $2\int sec^3 \, du$ use the Integration by parts formula and $2\int \sec u\, du= \ln|\sec u+\tan u|+C$

Answer 5

Let us consider the integral $$I=\int\sqrt{\frac{x-a}{x-b}} dx$$ and, to get rid as soon as possible of the square root, make a change of variable such as $$\frac{x-a}{x-b}=y^2$$ that is to say $$x=\frac{b y^2-a}{y^2-1}$$ which gives $$dx=\frac{2 y (a-b)}{\left(y^2-1\right)^2}dy$$ Back to the integral, we then have now $$I=(a-b)\int \frac{2 y^2 }{\left(y^2-1\right)^2} dy$$ The integrand can now be decomposed using partial fractions and $$\frac{2 y^2 }{\left(y^2-1\right)^2}=-\frac{1}{2 (y+1)}+\frac{1}{2 (y+1)^2}+\frac{1}{2 (y-1)}+\frac{1}{2 (y-1)^2}$$ which involves simple integrals and, as a final result $$\int \frac{2 y^2 }{\left(y^2-1\right)^2} dy=\frac{y}{1-y^2}+\frac{1}{2} \log (1-y)-\frac{1}{2} \log (1+y)=\frac{y}{1-y^2}-\tanh ^{-1}(y)$$

I am sure that you can take from here.