Trigonometric substitution and Integration of $\frac{1}{x^2\sqrt{x^2+1}} $ Regarding the integral
$$
\int \frac{dx}{x^2\sqrt{x^2 + 1}}
$$
I'm not sure what to do about the extra $x^2$ in the denominator. What can I do about it? 
 A: It actually makes your life easier. Make the usual substitution, $x=\tan\theta$. Then $dx=\sec^2\theta\,d\theta$, and your integral becomes
$$\int\frac{\sec^2\theta}{\tan^2\theta\sec\theta}d\theta=\int\frac{\sec\theta}{\tan^2\theta}d\theta\;.$$
Now convert everything to sines and cosines.
A: Setting $x=\tan\theta$, we have that $dx = \sec^2\theta \,d\theta$. Our denominator becomes $\tan^2\theta\sec\theta$ so then we want to evaluate
$$\int \frac{1}{\tan^2\theta \sec\theta}\sec^2\theta\,d\theta = \int \frac{\sec\theta}{\tan^2\theta}\,d\theta = \int\csc\theta\cot\theta\,d\theta$$
Can you take it from here? (Remember to change variables from $\theta$ back to $x$ after evaluating the integral.)
A: You might have to use the law of tangent which states that 
$$ \sec^2{\theta} =\tan^2{\theta} + 1  $$
or 
$$ \sec{\theta} = \sqrt{\tan^2{\theta} + 1}  $$
so 
$$ \int \frac{dx}{x^2\sqrt{x^2 + 1}}d\theta = \int \frac{d \tan \theta}{\tan^2\sqrt{\tan^2 + 1}}d\theta$$
therefore plugging in sec 
$$ \int \frac{d \tan \theta}{\tan^2\sqrt{\tan^2 + 1}}d\theta = \int \frac{\sec^2 \theta}{\tan^2\sec \theta}d\theta = \int \frac{\sec \theta}{\tan^2}d\theta $$
Therefore we now get answer
$$ \int \frac{\sec \theta}{\tan^2}d\theta = \int \csc \theta \cot \theta d\theta$$
Final answer 
$$ \int \csc \theta \cot \theta d\theta = -\csc \theta + C$$ 
because
$$ \frac{d}{dx} \csc \theta = -\csc \theta \cot \theta $$
A: $x=\cos \theta \implies dx=-\sin \theta\ d\theta$
$\therefore\displaystyle\int\dfrac{dx}{x^2\sqrt{x^2+1}}\\=-\displaystyle\int\dfrac{\sin \theta\ d\theta}{\cos^2 \theta\sqrt{\cos^2 \theta+1}}\\=-\displaystyle\int\dfrac{\sin \theta\ d\theta}{\cos^2 \theta\sqrt{2\cos^2 \left(\dfrac{\theta}{2}\right)}}\\=-\sqrt{2}\displaystyle\int\dfrac{\sin \left(\dfrac{\theta}{2}\right)\ d\theta}{2\cos^2\left(\dfrac{\theta}{2}\right)-1}\\$
$\cos\left(\dfrac{\theta}{2}\right)=z \implies-\dfrac{1}{2}\sin \left(\dfrac{\theta}{2}\right)d\theta=dz$ 
$\therefore-\sqrt{2}\displaystyle\int\dfrac{\sin \left(\dfrac{\theta}{2}\right)\ d\theta}{2\cos^2\left(\dfrac{\theta}{2}\right)-1}=\dfrac{1}{\sqrt{2}}\displaystyle\int\dfrac{dz}{2z^2-1}$
A: There's a way to this without trig substitutions
$$\int \frac{dx}{x^2\sqrt{1+x^2}} = \int \frac{dx}{x^3 \sqrt{\frac{1}{x^2}+1}}$$
Substitute $u = \frac{1}{x^2} + 1$, the integral becomes
$$ -\int \frac{du}{\sqrt{u}} $$
