How do you evaluate the integral $\int \frac{d x}{x^{2} \sqrt{x^{2} - 3}}$ $int dx/(x^2sqrt(x^2-3))$ ?

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How do you evaluate the integral $\int \frac{d x}{x^{2} \sqrt{x^{2} - 3}}$ $int dx/(x^2sqrt(x^2-3))$ ?

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Answer 1 · 2017-08-31T17:51:17

After using $x = \sqrt{3} \cdot sec u$ $x=sqrt(3)*secu$ and $d x = \sqrt{3} \cdot sec u \cdot tan u \cdot d u$ $dx=sqrt(3)*secu*tanu*du$ transforms, this integral became,

$\int \frac{\sqrt{3} \cdot sec u \cdot tan u \cdot (d u)}{3 {(sec u)}^{2} \cdot \sqrt{3} \cdot tan u}$ $int (sqrt(3)*secu*tanu*(du))/[3(secu)^2*sqrt(3)*tanu]$

= $\frac{1}{3} \cdot \int \frac{d u}{sec u}$ $1/3*int (du)/secu$

= $\frac{1}{3} \cdot \int cos u \cdot d u$ $1/3*int cosu*du$

= $\frac{1}{3} \cdot sin u + C$ $1/3*sinu+C$

After using $x = \sqrt{3} \cdot sec u$ $x=sqrt(3)*secu$ , $sec u = \frac{x}{\sqrt{3}}$ $secu=x/sqrt(3)$ , $tan u = \frac{\sqrt{x^{2} - 3}}{\sqrt{3}}$ $tanu=sqrt(x^2-3)/sqrt(3)$ and $sin u = \frac{tan u}{sec u} = \frac{\sqrt{x^{2} - 3}}{x}$ $sinu=tanu/secu=sqrt(x^2-3)/x$ inverse transforms, I found,

$\int \frac{d x}{x^{2} \cdot \sqrt{x^{2} - 3}} = \frac{\sqrt{x^{2} - 3}}{3 x} + C$ $int (dx)/[x^2*sqrt(x^2-3)]=sqrt(x^2-3)/(3x)+C$

Explanation:

I used $x = \sqrt{3} \cdot sec u$ $x=sqrt(3)*secu$ and $d x = \sqrt{3} \cdot sec u \cdot tan u \cdot d u$ $dx=sqrt(3)*secu*tanu*du$ transform

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How do you evaluate the integral $\int \frac{d x}{x^{2} \sqrt{x^{2} - 3}}$ $int dx/(x^2sqrt(x^2-3))$ ?

1 Answer

Explanation:

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How do you evaluate the integral $\int \frac{d x}{x^{2} \sqrt{x^{2} - 3}}$ $int dx/(x^2sqrt(x^2-3))$ ?