Question

\(\displaystyle \int \frac{dx}{1+36x^{2}}=\) ?

• A. \(6\tan^{-1}(6x)+k\)
• B. \(3\tan^{-1}(6x)+k\)
• C. \(\dfrac{1}{6}\tan^{-1}(6x)+k\)
• D. \(\tan^{-1}(6x)+k\)

Hint:

Standard form: $\displaystyle \int\frac{dx}{a^{2}+x^{2}}=\frac{1}{a}\tan^{-1}\!\frac{x}{a}+C$.

Answer 1

The Correct Option is C

Let \(u=6x\Rightarrow du=6\,dx\Rightarrow dx=\dfrac{du}{6}\). Then \[ \int \frac{dx}{1+(6x)^{2}}=\frac{1}{6}\int \frac{du}{1+u^{2}} =\frac{1}{6}\tan^{-1}u+k=\frac{1}{6}\tan^{-1}(6x)+k. \]