Question #fcb11

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Question #fcb11

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Answer 1 · 2017-08-31T03:29:08

You are correct: $\int \frac{\sqrt{x^{2} - 1}}{x^{2}} d x = ln ∣ ∣ x + \sqrt{x^{2} - 1} ∣ ∣ - \frac{\sqrt{x^{2} - 1}}{x} + C$ $intsqrt(x^2-1)/x^2dx=lnabs(x+sqrt(x^2-1))-sqrt(x^2-1)/x+C$

Explanation:

$I = \int \frac{\sqrt{x^{2} - 1}}{x^{2}} d x$ $I=intsqrt(x^2-1)/x^2dx$

I'm sure there are other methods of approach, but when I see square roots I immediately think of trig substitutions. Let's try $x = sec θ$ $x=sectheta$ , implying that $x^{2} - 1 = {sec}^{2} θ - 1 = {tan}^{2} θ$ $x^2-1=sec^2theta-1=tan^2theta$ and that $d x = sec θ tan θ d θ$ $dx=secthetatanthetad theta$ :

$I = \int \frac{\sqrt{{tan}^{2} θ}}{{sec}^{2} θ} (sec θ tan θ d θ)$ $I=intsqrt(tan^2theta)/sec^2theta(secthetatanthetad theta)$

$I = \int \frac{{tan}^{2} θ}{sec θ} d θ$ $I=inttan^2theta/secthetad theta$

$I = \int \frac{{sec}^{2} θ - 1}{sec θ} d θ$ $I=int(sec^2theta-1)/secthetad theta$

$I = \int (sec θ - cos θ) d θ$ $I=int(sectheta-costheta)d theta$

$I = ln | sec θ + tan θ | - sin θ + C$ $I=lnabs(sectheta+tantheta)-sintheta+C$

The substitution $x = sec θ$ $x=sectheta$ implies the following:

$tan θ = \sqrt{{sec}^{2} θ - 1} = \sqrt{x^{2} - 1}$ $tantheta=sqrt(sec^2theta-1)=sqrt(x^2-1)$
$cos θ = \frac{1}{x}$ $costheta=1/x$
$sin θ = \sqrt{1 - {cos}^{2} θ} = \sqrt{1 - \frac{1}{x^{2}}} = \frac{\sqrt{x^{2} - 1}}{x}$ $sintheta=sqrt(1-cos^2theta)=sqrt(1-1/x^2)=sqrt(x^2-1)/x$

Thus:

$I = ln ∣ ∣ x + \sqrt{x^{2} - 1} ∣ ∣ - \frac{\sqrt{x^{2} - 1}}{x} + C$ $I=lnabs(x+sqrt(x^2-1))-sqrt(x^2-1)/x+C$

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Question #fcb11

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