Trigonometric substitution and Integration of $\frac{1}{x^2\sqrt{x^2+1}} $

Question

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?

Answer 1

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.

Answer 2

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.)

Answer 3

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 $$ \frac{d}{dx} \csc \theta = -\csc \theta \cot \theta $$

Answer 4

$x=\cos \theta \implies dx=-\sin \theta\ d\theta$

$\therefore\displaystyle\int\dfrac{dx}{x^2\sqrt{x^2+1}}\\=-\displaystyle\int\dfrac{\sin \theta\ d\theta}{\cos^2 \theta\sqrt{\cos^2 \theta+1}}\\=-\displaystyle\int\dfrac{\sin \theta\ d\theta}{\cos^2 \theta\sqrt{2\cos^2 \left(\dfrac{\theta}{2}\right)}}\\=-\sqrt{2}\displaystyle\int\dfrac{\sin \left(\dfrac{\theta}{2}\right)\ d\theta}{2\cos^2\left(\dfrac{\theta}{2}\right)-1}\\$

$\cos\left(\dfrac{\theta}{2}\right)=z \implies-\dfrac{1}{2}\sin \left(\dfrac{\theta}{2}\right)d\theta=dz$

$\therefore-\sqrt{2}\displaystyle\int\dfrac{\sin \left(\dfrac{\theta}{2}\right)\ d\theta}{2\cos^2\left(\dfrac{\theta}{2}\right)-1}=\dfrac{1}{\sqrt{2}}\displaystyle\int\dfrac{dz}{2z^2-1}$

Answer 5

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

$$ -\int \frac{du}{\sqrt{u}} $$