Trigonometric Substitution on $\frac{1}{x\sqrt{(x^2 +25)}}$ How can I find
$$\int\frac{1}{x\sqrt{(x^2 +25)}} \space dx$$
using trigonometric substitution?
 A: $$\int \frac{1}{x\sqrt{(x^2+25)}} \space \text{d}x =$$

Substitute $x=5\tan(u)$ and $\text{d}x=5\sec^2(u)\text{d}u$. Then $\sqrt{x^2+25}=\sqrt{25\tan^2(u)+25}=5\sec(u)$ and $u=\tan^{-1}\left(\frac{x}{5}\right)$:

$$5 \int \frac{\csc(u)}{25} \space \text{d}u =$$
$$\frac{5}{25} \int \csc(u) \space \text{d}u =$$
$$\frac{1}{5} \int \csc(u) \space \text{d}u =$$
$$\frac{1}{5} \int -\frac{-\cot(u)\csc(u)-\csc^2(u)}{\cot(u)+\csc(u)} \space \text{d}u =$$

Substitute $s=\cot(u)+\csc(u)$ and $\text{d}s=(-\csc^2(u)-\cot(u)\csc(u))\text{d}u$:

$$\frac{1}{5} \int -\frac{1}{s} \space \text{d}s =$$
$$-\frac{1}{5} \int \frac{1}{s} \space \text{d}s =$$
$$-\frac{1}{5} \cdot \ln\left(s\right) +C =$$
$$-\frac{\ln\left(\cot(u)+\csc(u)\right)}{5}+C =$$
$$-\frac{\ln\left(\cot\left(\tan^{-1}\left(\frac{x}{5}\right)\right)+\csc\left(\tan^{-1}\left(\frac{x}{5}\right)\right)\right)}{5}+C =$$
$$-\frac{\ln\left(\frac{5+\sqrt{x^2+25}}{x}\right)}{5}+C$$
A: You can solve this one using this substitution: $x = 5\tan{u}$. Then $dx = 5\sec^2{(u)}du$, and we can rewrite the integral as $\int{\frac{5\sec^2{(u)}du}{(5\tan{u} )\sqrt{25\tan^2{(u)}+25}}}$. We can use one of the Pythagorean Identities to rewrite the radical as $\sqrt{25\tan^2{(u)}+25}=5\sec{u}$, and we're left with $\int{\frac{\sec{u}}{5\tan{u}}du} = \int{\frac{du}{5\sin{u}}}$. 
You can use a second round of $u$-substitution to integrate this one, or you can use the $\csc{u}$ integration rule, to get: $\int{\frac{\csc{u}}{5}du} = \frac{1}{5}\ln|\csc{u}-\cot{u}|+C$.
We just need to substitute back for $u$, which is $\tan^{-1}{\frac{x}{5}}$, to get the indefinite integral in terms of $x$:
$\frac{1}{5}\ln|\csc{\tan^{-1}{\frac{x}{5}}}-\cot{\tan^{-1}{\frac{x}{5}}}|+C$.
Note: It is possible to convert this to a non-trig form using $\log$s and radicals. 
A: Looks like the trig substitution is the best approach but it leads to an integral that is difficult to evaluate in itself: 
substitute $x=5\tan(\theta)$. Then $\sqrt{(x^2+25)}=5\sec(\theta)$ and $dx=5\sec^2(\theta)d\theta.$ Simplifying the integral becomes: $\int{csc(\theta)d\theta}$. Any ideas about the best way to evaluate this integral?
