Question

ok so =tacrtan4t-
∫4t/(1+16t2)dt
i believe I need to modify this last integral further so that I my final answer has a ln component. How do I do that?

Answer 1

Substitute \(1+16t^2=u\).

Answer 2

ok got it. Thanks!!!

Answer 3

Or rewrite the integral as,
\(\frac{1}{8}\int\limits \frac{8(4t)}{1+16t^2}dt. \)