\(\int\frac{1}{x^2+16}dx=\)
Answer & explanation
Correct answer: option 3
\(\int\frac{1}{x^2+16}dx=\int\frac{1}{x^2+(4)^2}dx\)
$=\frac{1}{4}\tan^{-1}\frac{x}{4}+c$
\(\int\frac{1}{x^2+16}dx=\)
Correct answer: option 3
\(\int\frac{1}{x^2+16}dx=\int\frac{1}{x^2+(4)^2}dx\)
$=\frac{1}{4}\tan^{-1}\frac{x}{4}+c$