Can anyone help me solve this?
Integral of √(4+x^2) / x^5 dx? Thanks.
This one's tricky. It requires both a trigonometric substitution and a $\displaystyle u$ substitution.
First, use the substitution $\displaystyle x = 2\tan{\theta}$, which means $\displaystyle dx = 2\sec^2{\theta}\,d\theta$.
Putting this into the integral...
$\displaystyle \int{\frac{\sqrt{4 + x^2}}{x^5}\,dx} = \int{\frac{\sqrt{4 + (2\tan{\theta})^2}}{(2\tan{\theta})^5}\cdot 2\sec^2{\theta}\,d\theta}$
$\displaystyle = \int{\frac{2\sec^2{\theta}\sqrt{4(1 + \tan^2{\theta})}}{32\tan^5{\theta}}\,d\theta}$
$\displaystyle = \int{\frac{2\sec^2{\theta}\sqrt{4\sec^2{\theta}}}{ 32\tan^5{\theta}}\,d\theta}$
$\displaystyle = \int{\frac{4\sec^4{\theta}}{32\tan^5{\theta}}\,d\t heta}$
$\displaystyle = \frac{1}{8}\int{\frac{\sec^4{\theta}}{\tan^5{\thet a}}\,d\theta}$
$\displaystyle = \frac{1}{8}\int{\frac{\cos{\theta}}{\sin^5{\theta} }\,d\theta}$.
Now we can use a $\displaystyle u$ substitution.
Let $\displaystyle u = \sin{\theta}$ so that $\displaystyle du = \cos{\theta}\,d\theta$.
So the integral becomes
$\displaystyle \frac{1}{8}\int{u^{-5}\,du}$
$\displaystyle = -\frac{1}{32}u^{-4} + C$
$\displaystyle = -\frac{1}{32}\sin^{-4}{\theta} + C$
Now, since $\displaystyle 1 + \cot^2{\theta} = \csc^2{\theta}$, we have
$\displaystyle 1 + \frac{1}{\tan^2{\theta}} = \sin^{-2}{\theta}$
$\displaystyle \left(1 + \frac{1}{\tan^2{\theta}}\right)^2 = \sin^{-4}{\theta}$.
Also remember that $\displaystyle x = 2\tan{\theta}$, so $\displaystyle \tan{\theta} = \frac{x}{2}$.
Thus $\displaystyle \sin^{-4}{\theta} = \left(1 + \frac{4}{x^2}\right)^2$.
So finally, the answer to the original integral is...
$\displaystyle -\frac{1}{32}\left(1 + \frac{4}{x^2}\right)^2 + C$.