Question

use the accompanying sections of a table of integrals to evaluate the following indefinite integral. the integral may require preliminary work, such as completing the square or changing variables, before it can be found in a table.
int\frac{sqrt{ln^{2}x + 16}}{x}dx
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int\frac{sqrt{ln^{2}x + 16}}{x}dx=square
(use parentheses to clearly denote the argument of each function.)

Explanation:

Step1: Use substitution

Let $u = \ln x$, then $du=\frac{1}{x}dx$. The integral $\int\frac{\sqrt{\ln^{2}x + 16}}{x}dx$ becomes $\int\sqrt{u^{2}+16}du$.

Step2: Recall integral formula

The integral of $\int\sqrt{u^{2}+a^{2}}du=\frac{u}{2}\sqrt{u^{2}+a^{2}}+\frac{a^{2}}{2}\ln(u + \sqrt{u^{2}+a^{2}})+C$. Here $a = 4$. So $\int\sqrt{u^{2}+16}du=\frac{u}{2}\sqrt{u^{2}+16}+8\ln(u+\sqrt{u^{2}+16})+C$.

Step3: Substitute back

Substitute $u=\ln x$ back into the result. We get $\frac{\ln x}{2}\sqrt{\ln^{2}x + 16}+8\ln(\ln x+\sqrt{\ln^{2}x + 16})+C$.

Answer:

$\frac{\ln x}{2}\sqrt{\ln^{2}x + 16}+8\ln(\ln x+\sqrt{\ln^{2}x + 16})+C$