Practicing Success
Practicing Success
CUET Preparation Today
\(\int\frac{1}{x^2+16}dx=\) |
\(\tan^{-1}4x+c\) \(\tan^{-1}\frac{x}{4}+c\) \(\frac{1}{4}\tan^{-1}\frac{x}{4}+c\) none |
\(\frac{1}{4}\tan^{-1}\frac{x}{4}+c\) |
