Question

∫ dx/(x^2+16)

My teacher has the answer as 1/4 tan^-1(1/4x) +C
I have no idea where the tan came from, there's no trig function in the problem.

Answer 1

\[\int\frac{1}{x^2+16}dx\]

\[=\int\frac{1}{16(\frac{x^2}{16}+1)}dx\]

\[=\frac{1}{16}\int\frac{1}{\left(\frac{x}{4}\right)^2+1}dx\]

Answer 2

So my teacher is incorrect?

Answer 3

No, the inverse tangent comes from a trigonometric substitution. Notice how the denominator is a lot like the Pythagorean identity (with some minor alterations)?
\[\int\frac{\color{red}{dx}}{\color{blue}{x}^2+16}\]
Let \(\color{blue}{x=4\tan u}\), so you also have \(\color{red}{dx=4\sec^2u~du}\).
\[\int\frac{\color{blue}{4\sec^2u}}{(\color{red}{4\tan u})^2+16}~\color{blue}{du}\]