Finding the integral $\int \sqrt{1-\frac{3}{x}+\frac{1}{x^2}} \, dx$

Question

How to find the integral $$\displaystyle\int \sqrt{1-\dfrac{3}{x}+\dfrac{1}{x^2}}\mathrm dx$$ ?

I cannot think of any suitable method. I tried trigonometric substitutions but they were of no use. Hints please!

Answer 1

Use the Euler substitution of the first kind, let $\sqrt{x^2-3x+1}=x+t$,we have \begin{align*} \int{\frac{\sqrt{x^2-3x+1}}{x}}\text{d}x&=-2\int{\frac{\left( t^2+3t+1 \right) ^2}{\left( 1-t^2 \right) \left( 2t+3 \right) ^2}}\text{d}t \\ &=-2\int\left ( \frac{1}{2}\frac{1}{t+1}-\frac{3}{2}\frac{1}{2t+3}-\frac{5}{4}\frac{1}{\left ( 2t+3\right )^{2}}-\frac{1}{2}\frac{1}{t-1}-\frac{1}{4}\right ) \mathrm{d}t \end{align*} then you can take it from here.

Answer 2

Integrate by parts to break up the integral \begin{align} &\int \sqrt{1-\frac{3}{x}+\frac{1}{x^2}} \, dx\\ =& \ x \sqrt{1-\frac{3}{x}+\frac{1}{x^2}} +\int \frac1{x^2\sqrt{1-\frac{3}{x}+\frac{1}{x^2}}}dx -\frac32 \int \frac1{x\sqrt{1-\frac{3}{x}+\frac{1}{x^2}}}dx \end{align} and then apply the Euler substitution $\sqrt{1-\frac{3}{x}+\frac{1}{x^2}}=t-\frac1x$ to evaluate \begin{align} &\int \frac1{x^2\sqrt{1-\frac{3}{x}+\frac{1}{x^2}}}dx =\int\frac2{3-2t}dt = -\ln|3-2t|\\ &\int \frac1{x\sqrt{1-\frac{3}{x}+\frac{1}{x^2}}}dx =\int\frac2{1-t^2}dt = -\ln |\frac{t-1}{t+1}|\\ \end{align}