How proof this answer is right ? ∫ ((1)/((x^2)(sqrt(1+x^2))dx =

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How proof this answer is right ? ∫ ((1)/((x^2)(sqrt(1+x^2))dx =

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Answer 1 · 2015-09-09T21:08:29

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Answer 2 · 2015-09-09T22:55:16

Show that the derivative of what you think is the answer is the integrand in the original problem.

Explanation:

Quick example:
To prove that $\int e^{3 x} d x = \frac{1}{3} e^{3 x} + C$ $inte^(3x) dx = 1/3 e^(3x)+C$ , show that the derivative of $\frac{1}{3} e^{3 x} + C$ $1/3 e^(3x)+C$ is $e^{3 x}$ $e^(3x)$ :

$\frac{1}{3} e^{3 x} \cdot 3 = e^{3 x}$ $1/3 e^(3x)*3 = e^(3x)$ . So the answer is correct.

In your question, show that the derivative of $- \frac{\sqrt{1 + x^{2}}}{x} + C$ $-sqrt(1+x^2)/x +C$ is $\frac{1}{x^{2} \sqrt{1 + x^{2}}}$ $1/(x^2sqrt(1+x^2))$

It is clear that we can ignore the $+ C$ $+C$ because the derivative of a constant is $0$ $0$ .

Answer 3 · 2015-09-10T01:08:50

You could do the actual integral and check each step along the way.

$\int \frac{1}{x^{2} \sqrt{1 + x^{2}}} d x = ?$ $int 1/(x^2sqrt(1+x^2))dx = ?$

Let:
$x = tan θ$ $x = tantheta$
$d x = {sec}^{2} θ d θ$ $dx = sec^2thetad theta$
$\sqrt{1 + x^{2}} = \sqrt{1 + {tan}^{2} θ} = sec θ$ $sqrt(1+x^2) = sqrt(1+tan^2theta) = sectheta$
$x^{2} = {tan}^{2} θ$ $x^2 = tan^2theta$

This is a typical trig substitution strategy and is verified in any calculus textbook that teaches this. You can check the $\sqrt{1 + x^{2}}$ $sqrt(1+x^2)$ substitution and you will see that it is really $\sqrt{1 + {tan}^{2} θ} = \sqrt{{sec}^{2} θ} = sec θ$ $sqrt(1+tan^2theta) = sqrt(sec^2theta) = sectheta$ .

Therefore you get:

$\Rightarrow \int \frac{1}{{tan}^{2} θ sec θ} {sec}^{2} θ d θ$ $=> int 1/(tan^2thetasectheta)sec^2thetad theta$

Using identities $tan θ = \frac{sin θ}{cos θ}$ $tantheta = sintheta/costheta$ and $sec θ = \frac{1}{cos θ}$ $sectheta = 1/costheta$ :
$= \int \frac{{cos}^{2} θ}{{sin}^{2} θ} \cdot \frac{1}{cos θ} d θ$ $= int cos^2theta/sin^2theta*1/costhetad theta$

$= \int \frac{cos θ}{{sin}^{2} θ} d θ$ $= int costheta/sin^2thetad theta$

Some u-substitution can be done now as well. Let:
$u = sin θ$ $u = sintheta$
$d u = cos θ d θ$ $du = costhetad theta$

$\Rightarrow \int \frac{1}{u^{2}} d u$ $=> int 1/u^2du$

$= - \frac{1}{u} = - \frac{1}{sin θ}$ $= -1/u = color(green)(-1/sintheta)$

That is our temporary answer. Finally, going back to the original substitution of $x = tan θ$ $x = tantheta$ :
$tan θ = \frac{sin θ}{cos θ} = x$ $tantheta = sintheta/costheta = x$

$sin θ = x cos θ$ $sintheta = xcostheta$

$sec θ = \sqrt{1 + x^{2}} \Rightarrow cos θ = \frac{1}{\sqrt{1 + x^{2}}}$ $sectheta = sqrt(1+x^2) => costheta = 1/sqrt(1+x^2)$

$sin θ = \frac{x}{\sqrt{1 + x^{2}}} \Rightarrow \frac{1}{sin θ} = \frac{\sqrt{1 + x^{2}}}{x}$ $sintheta = x/sqrt(1+x^2) => 1/sintheta = sqrt(1+x^2)/x$

Therefore the final answer is indeed:

$= - \frac{1}{sin θ} + C = - \frac{\sqrt{1 + x^{2}}}{x} + C$ $= -1/sintheta + C = color(blue)(-sqrt(1+x^2)/x + C)$

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How proof this answer is right ? ∫ ((1)/((x^2)(sqrt(1+x^2))dx =

3 Answers

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