∫116+x2 dx\int \frac{1}{16+{x}^{2}} \, dx
1
Use Trigonometric Substitution
Let x=4tan⁡ux=4\tan{u}, dx=4sec⁡2u dudx=4\sec^{2}u \, du

2
Substitute variables from above.
∫116+(4tan⁡u)2×4sec⁡2u du\int \frac{1}{16+{(4\tan{u})}^{2}}\times 4\sec^{2}u \, du

3
Simplify.
∫14 du\int \frac{1}{4} \, du

4
Use this rule: ∫a dx=ax+C\int a \, dx=ax+C.
u4\frac{u}{4}

5
From the earlier steps, we know that:
u=tan⁡−1(14x)u=\tan^{-1}{(\frac{1}{4}x)}

6
Substitute the above back into the original integral.
tan⁡−1(14x)4\frac{\tan^{-1}{(\frac{1}{4}x)}}{4}

7
Add constant.
tan⁡−1(x4)4+C\frac{\tan^{-1}{(\frac{x}{4})}}{4}+C

Done
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