How do you integrate $\int \frac{1}{x^{2} \sqrt{x^{2} - 7}}$ $int 1/(x^2sqrt(x^2-7))$ by trigonometric substitution?

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How do you integrate $\int \frac{1}{x^{2} \sqrt{x^{2} - 7}}$ $int 1/(x^2sqrt(x^2-7))$ by trigonometric substitution?

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Answer 1 · 2016-11-25T22:23:43

$\frac{\sqrt{x^{2} - 7}}{7 x} + C$ $sqrt(x^2-7)/(7x)+C$

Explanation:

$I = \int \frac{d x}{x^{2} \sqrt{x^{2} - 7}}$ $I=intdx/(x^2sqrt(x^2-7))$

Apply the substitution $x = \sqrt{7} sec θ, d x = \sqrt{7} sec θ tan θ d θ$ $x=sqrt7sectheta,dx=sqrt7secthetatanthetad theta$ .

$I = \int \frac{\sqrt{7} sec θ tan θ d θ}{7 {sec}^{2} θ \sqrt{7 {sec}^{2} θ - 7}}$ $color(white)I=int(sqrt7secthetatanthetad theta)/(7sec^2thetasqrt(7sec^2theta-7))$

$I = \frac{1}{7} \int \frac{tan θ d θ}{sec θ \sqrt{{sec}^{2} θ - 1}}$ $color(white)I=1/7int(tanthetad theta)/(secthetasqrt(sec^2theta-1))$

$I = \frac{1}{7} \int \frac{d θ}{sec θ}$ $color(white)I=1/7int(d theta)/sectheta$

$I = \frac{1}{7} \int cos θ d θ$ $color(white)I=1/7intcosthetad theta$

$I = \frac{1}{7} sin θ$ $color(white)I=1/7sintheta$

$I = \frac{1}{7} \frac{tan θ}{sec θ}$ $color(white)I=1/7tantheta/sectheta$

$I = \frac{1}{7} \frac{\sqrt{{sec}^{2} θ - 1}}{sec θ}$ $color(white)I=1/7sqrt(sec^2theta-1)/sectheta$

Using $x = \sqrt{7} sec θ \Rightarrow sec θ = \frac{x}{\sqrt{7}}$ $x=sqrt7sectheta=>sectheta=x/sqrt7$ :

$I = \frac{1}{7} \frac{\sqrt{\frac{x^{2}}{7} - 1}}{\frac{x}{\sqrt{7}}}$ $color(white)I=1/7sqrt(x^2/7-1)/(x/sqrt7)$

$I = \frac{1}{7} \frac{\frac{1}{\sqrt{7}} \sqrt{x^{2} - 7}}{\frac{x}{\sqrt{7}}}$ $color(white)I=1/7(1/sqrt7sqrt(x^2-7))/(x/sqrt7)$

$I = \frac{\sqrt{x^{2} - 7}}{7 x} + C$ $color(white)I=sqrt(x^2-7)/(7x)+C$

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How do you integrate $\int \frac{1}{x^{2} \sqrt{x^{2} - 7}}$ $int 1/(x^2sqrt(x^2-7))$ by trigonometric substitution?

1 Answer

Explanation:

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How do you integrate int 1/(x^2sqrt(x^2-7))∫1x2√x2−7int 1/(x^2sqrt(x^2-7)) by trigonometric substitution?

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How do you integrate $\int \frac{1}{x^{2} \sqrt{x^{2} - 7}}$ $int 1/(x^2sqrt(x^2-7))$ by trigonometric substitution?