<s>[INST] 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? 
 [/INST] 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
$$ \frac{d}{dx} \csc \theta = -\csc \theta \cot \theta $$
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