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

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!

• HAve you tried $\int\frac{\sqrt{x^2-3x+1}}{x}\,dx$? – Tito Eliatron Jan 18 '17 at 10:39
• @TitoEliatron Isn't that the same thing? What should I try with it ? – user404484 Jan 18 '17 at 10:40
• Yeah! $x^2-3x+1=(x-3/2)^2-5/4$ and you can use the C.V. $x-3/2=\sqrt5/2 \sec t$ or $\sqrt{x^2-3x+1}=x+t$ – Tito Eliatron Jan 18 '17 at 10:44
• @TitoEliatron I used the first substitution and I'm getting $$\int \frac{\tan(t)}{\sec(t)+\frac{3}{\sqrt(5)}}\sec(t)\tan(t)dt$$ What to do next? – user404484 Jan 18 '17 at 10:56

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.