<s>[INST] 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?
 [/INST] $$\int \frac{1}{x\sqrt{(x^2+25)}} \space \text{d}x =$$

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

$$\frac{1}{5} \int -\frac{1}{s} \space \text{d}s =$$
$$-\frac{1}{5} \int \frac{1}{s} \space \text{d}s =$$
$$-\frac{1}{5} \cdot \ln\left(s\right) +C =$$
$$-\frac{\ln\left(\cot(u)+\csc(u)\right)}{5}+C =$$
$$-\frac{\ln\left(\cot\left(\tan^{-1}\left(\frac{x}{5}\right)\right)+\csc\left(\tan^{-1}\left(\frac{x}{5}\right)\right)\right)}{5}+C =$$
$$-\frac{\ln\left(\frac{5+\sqrt{x^2+25}}{x}\right)}{5}+C$$
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