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

2
Substitute variables from above.
∫(4sec⁡u)2−164sec⁡u×4sec⁡utan⁡u du\int \frac{\sqrt{{(4\sec{u})}^{2}-16}}{4\sec{u}}\times 4\sec{u}\tan{u} \, du

3
Simplify.
∫4tan⁡2u du\int 4\tan^{2}u \, du

4
Use Constant Factor Rule: ∫cf(x) dx=c∫f(x) dx\int cf(x) \, dx=c\int f(x) \, dx.
4∫tan⁡2u du4\int \tan^{2}u \, du

5
Use Pythagorean Identities: tan⁡2x=sec⁡2x−1\tan^{2}x=\sec^{2}x-1.
4∫sec⁡2u−1 du4\int \sec^{2}u-1 \, du

6
Use Sum Rule: ∫f(x)+g(x) dx=∫f(x) dx+∫g(x) dx\int f(x)+g(x) \, dx=\int f(x) \, dx+\int g(x) \, dx.
4(∫sec⁡2u du+∫−1 du)4(\int \sec^{2}u \, du+\int -1 \, du)

7
The derivative of tan⁡u\tan{u} is sec⁡2u\sec^{2}u.
4(tan⁡u+∫−1 du)4(\tan{u}+\int -1 \, du)

8
Use this rule: ∫a dx=ax+C\int a \, dx=ax+C.
4(tan⁡u−u)4(\tan{u}-u)

9
From the earlier steps, we know that:
sec⁡u=14xtan⁡u=(14x)2−1\begin{aligned}&\sec{u}=\frac{1}{4}x\\&\tan{u}=\sqrt{{(\frac{1}{4}x)}^{2}-1}\end{aligned}

10
Substitute the above back into the original integral.
4((14x)2−1−sec⁡−1(14x))4(\sqrt{{(\frac{1}{4}x)}^{2}-1}-\sec^{-1}{(\frac{1}{4}x)})

11
Add constant.
4(x216−1−sec⁡−1(x4))+C4(\sqrt{\frac{{x}^{2}}{16}-1}-\sec^{-1}{(\frac{x}{4})})+C

Done
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