∫ 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.

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.

zarkon · Answer

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

anonymous · Answer

So my teacher is incorrect?

anonymous · Answer

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 $\color{red}{dx=4\sec^2u~du}$.
$$\int\frac{\color{blue}{4\sec^2u}}{(\color{red}{4\tan u})^2+16}~\color{blue}{du}$$