Question

Integrate the following

Answer 1

I suppose you mean to have the entire numerator under the square root?

`\displaystyle\int_2^4(√(x^2-4))/(x^2)\,\mathrm dx`

We can use a trigonometric substitution to start:

`x=2\sec t\implies\mathrm dx=2\sec t\tan t\,\mathrm dt`

Then for
`x=2`,
`t=\sec^(-1)1=0`; for
`x=4`,
`t=\sec^(-1)2=\frac\pi3`. So the integral is equivalent to

`\displaystyle\int_0^(\pi/3)(√((2\sec t)^2-4))/((2\sec t)^2)(2\sec t\tan t)\,\mathrm dt=\int_0^(\pi/3)(\tan^2t)/(\sec t)\,\mathrm dt`

We can write

`(\tan^2t)/(\sec t)=\frac{(\sin^2t)/(\cos^2t)}{\frac1{\cos t}}=(\sin^2t)/(\cos t)=(1-\cos^2t)/(\cos t)=\sec t-\cos t`

so the integral becomes

`\displaystyle\int_0^(\pi/3)(\sec t-\cos t)\,\mathrm dt=\boxed{\ln(2+\sqrt3)-\frac{\sqrt3}2}`