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

Sep 9, 2015

#### Explanation:

Clear image of the equation

Sep 9, 2015

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

#### Explanation:

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

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

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

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

Sep 10, 2015

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

int 1/(x^2sqrt(1+x^2))dx = ?

Let:
$x = \tan \theta$
$\mathrm{dx} = {\sec}^{2} \theta d \theta$
$\sqrt{1 + {x}^{2}} = \sqrt{1 + {\tan}^{2} \theta} = \sec \theta$
${x}^{2} = {\tan}^{2} \theta$

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

Therefore you get:

$\implies \int \frac{1}{{\tan}^{2} \theta \sec \theta} {\sec}^{2} \theta d \theta$

Using identities $\tan \theta = \sin \frac{\theta}{\cos} \theta$ and $\sec \theta = \frac{1}{\cos} \theta$:
$= \int {\cos}^{2} \frac{\theta}{\sin} ^ 2 \theta \cdot \frac{1}{\cos} \theta d \theta$

$= \int \cos \frac{\theta}{\sin} ^ 2 \theta d \theta$

Some u-substitution can be done now as well. Let:
$u = \sin \theta$
$\mathrm{du} = \cos \theta d \theta$

$\implies \int \frac{1}{u} ^ 2 \mathrm{du}$

$= - \frac{1}{u} = \textcolor{g r e e n}{- \frac{1}{\sin} \theta}$

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

$\sin \theta = x \cos \theta$

$\sec \theta = \sqrt{1 + {x}^{2}} \implies \cos \theta = \frac{1}{\sqrt{1 + {x}^{2}}}$

$\sin \theta = \frac{x}{\sqrt{1 + {x}^{2}}} \implies \frac{1}{\sin} \theta = \frac{\sqrt{1 + {x}^{2}}}{x}$

Therefore the final answer is indeed:

$= - \frac{1}{\sin} \theta + C = \textcolor{b l u e}{- \frac{\sqrt{1 + {x}^{2}}}{x} + C}$