Question

the integral of 1/(16x^2+1)

Answer 1

Let \(x=\dfrac{1}{4}\tan u\), so \(dx=\dfrac{1}{4}\sec^2u~du\).
\[\int\frac{1}{16x^2+1}~dx=\int\frac{\dfrac{1}{4}\sec^2u}{16\left(\dfrac{1}{4}\tan u\right)^2+1}~du=\frac{1}{4}\int\frac{\sec^2u}{\tan^2u+1}~du\]

Answer 2

ok why does x =1/4tanu