How to evaluate $\int\frac{\sqrt{x^2 - 4}}{x^3}$?

Question

I've tried to substitute $x = 2 \cosh(t)$ but it ended up: $$\frac 12\int(\operatorname{sech}t-\operatorname{sech}^3 t)\,dt$$

I can solve $\int\operatorname{sech} t\,dt$ but $\int\operatorname{sech}^3 t\,dt$ really kills me.

Or maybe I'm so frustrated I can't continue.

Answer 1

set $\sqrt{x^2-4}=y, x^2=y^2+4, x\ dx=y\ dy$

$$I=\int\dfrac{\sqrt{x^2-4}\cdot x\ dx}{x^4}=\int\dfrac{y^2\ dy}{(y^2+4)^2}$$

Now integrate by parts $$I=y\int\dfrac{y\ dy}{(y^2+4)^2}-\int\left(\dfrac{dy}{dy}\int\dfrac{y\ dy}{(y^2+4)^2}\right)dy$$

Alternatively $x=2\sec t$

$$I=\int\dfrac{2\tan t}{(2\sec t)^3}\cdot2\sec t\tan t\ dt=\dfrac12\int\sin^2tdt=\dfrac14\int(1-\cos2t)dt=?$$

Answer 2

Integration by parts gives $$I=\int \text{sech} x ~\text{sech}^2~ x ~dx=\text{sech} x \tanh x+\int \text{sech} x \$$ tanh^2 x ~dx$$ $$I=\text{sech} x \tanh x+\int \text{sech} x (1-\text{sech}^2 x) dx$$ $$\implies 2I=\text{sech}x \tanh x+\int \text{sech}~x dx \implies I=\frac{1}{2}\text{sech}x \tanh x+\tan^{-1}\tanh(x/2)+C$$

And use$$\int \text{sech}x ~dx=2\tan^{-1} \tanh(x/2)+D$$

Answer 3

Set $\frac{2}{x}=y$, so $-\frac{2}{x^2}\text{d}x=\text{d}y$; hence $$\int \frac{\sqrt{x^2-4}}{x^3} \text{d}x=-\frac{1}{4} \int y \sqrt{\frac{4}{y^2}-4} \text{d}y=-\frac{1}{2}\ \int \sqrt{1-y^2} \text{d}y$$ Now set $y=\sin z$, so $\text{d}y=\cos z \text{d}z$; so we have $$-\frac{1}{2} \int \cos^2 z \text{d}z$$ And this one can be easily evaluated using $\cos^2 z = \frac{\cos(2z)+1}{2}$.

Answer 4

Hint. Another way to do this is to set $x=2\sec\phi,$ so that the integral becomes $$\int \frac{2\tan\phi}{8\sec^3\phi}2\sec\phi\tan\phi\mathrm d\phi,$$ or in other words, $$\frac12\int \sin^2\phi\mathrm d\phi,$$ from where you should be able to continue.