Question

Evaluate the indefinite integral. (Use C for the constant of integration.)

Answer 1

\[\int\limits_{}^{}\frac{ \tan^{-1}3x }{ 1+9x^2 }dx\]

Answer 2

\[Let u = \tan^{-1}3x, du = \frac{ 3 }{ 1+9x^2 }\]
\[\int\limits_{}^{}\tan^{-1}3x * \frac{ 1 }{ 1+9x^2 }dx = \int\limits_{}^{}udu\]
What I have so far

Answer 3

du = 3/(1+9x^2)

so

du/3 = 1/(1+9x^2)

Answer 4

Okay, got it now, missed that tiny detail of (1/3)du = 1/(1+9x^2).
Answer: \[\frac{ 1 }{ 6 }(\tan^{-1}3x)^2+C\]

Answer 5

correct