Question

Integral of 1 /x+ sq.root x^2-1

Answer 1

I'm assuming the integral is

$\displaystyle \int \frac{\mathrm dx}{x+\sqrt{x^2-1}}$

Rationalize the denominator:

$\dfrac1{x+\sqrt{x^2-1}} \times \dfrac{x-\sqrt{x^2-1}}{x-\sqrt{x^2-1}} = \dfrac{x-\sqrt{x^2-1}}{x^2-\left(\sqrt{x^2-1}\right)^2} = x-\sqrt{x^2-1}$

Then the integral is

$\displaystyle \int\left(x-\sqrt{x^2-1}\right)\,\mathrm dx = \dfrac12x^2 - \int\sqrt{x^2-1}\,\mathrm dx$

For the remaining integral, substitute x = sec(t ) and dx = sec(t ) tan(t ) dt. Then over an appropriate domain, we have

$\displaystyle\int\sqrt{x^2-1}\,\mathrm dx = \int\sec(t)\tan(t)\sqrt{\sec^2(t)-1}\,\mathrm dt = \int\sec(t)\tan^2(t)\,\mathrm dt$

Integrate by parts, taking

u = tan(t )   ==>   du = sec²(t ) dt

dv = sec(t ) tan(t ) dt   ==>   v = sec(t )

Then

$\displaystyle\int\sec(t)\tan^2(t)\,\mathrm dt = \sec(t)\tan(t) - \int\sec^3(t)\,\mathrm dt$

Now for *this* remaining integral, integrate by parts again, taking

u = sec(t )   ==>   du = sec(t ) tan(t ) dt

dv = sec²(t ) dt   ==>   v = tan(t )

so that

$\displaystyle\int\sec^3(t)\,\mathrm dt = \sec(t)\tan(t) - \int\sec(t)\tan^2(t)\,\mathrm dt \\\\ \int\sec^3(t)\,\mathrm dt = \sec(t)\tan(t) - \int\sec(t)(\sec^2(t)-1)\,\mathrm dt \\\\ \int\sec^3(t)\,\mathrm dt = \sec(t)\tan(t) - \int\sec^3(t)\,\mathrm dt + \int\sec(t)\,\mathrm dt \\\\ \int\sec^3(t)\,\mathrm dt = \frac12\sec(t)\tan(t)+\frac12\int\sec(t)\,\mathrm dt \\\\ \int\sec^3(t)\,\mathrm dt = \frac12\sec(t)\tan(t) + \frac12\ln\left|\sec(t)+\tan(t)\right| + C$

To summarize, if I denotes the original integral, we have

$\displaystyle I = \frac12x^2 - \int\sqrt{x^2-1}\,\mathrm dx \\\\ I = \frac12x^2 - \int\sec(t)\tan^2(t)\,\mathrm dt \\\\ I = \frac12x^2 - \sec(t)\tan(t) + \int\sec^3(t)\,\mathrm dt \\\\ I = \frac12x^2 - \sec(t)\tan(t) + \frac12\sec(t)\tan(t) + \frac12\ln\left|\sec(t)+\tan(t)\right| + C \\\\ I = \frac12x^2 - \frac12\sec(t)\tan(t) + \frac12\ln\left|\sec(t)+\tan(t)\right| + C$

Putting everything back in terms of x, we have

sec(t ) = x

tan(t ) = √(x ² - 1)

so that

$\displaystyle I = \boxed{\frac12x^2 - \frac12x\sqrt{x^2-1}+\frac12\ln\left|x+\sqrt{x^2-1}\right|+C}$