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tandoorichicken
Oct10-04, 06:59 PM
How does one integrate
\int\arctan 4t\,dt
?

This is out of a section on integration by parts. Maybe that would help but I don't see any "parts." I only see one, the arctan.

Math Is Hard
Oct10-04, 07:03 PM
Let u = arctan4t
dv = dt

tandoorichicken
Oct10-04, 07:20 PM
Thank you so much. Please tell me if I did this correctly.
Integration by Parts:
\int u\,dv = uv - \int v\,du

u = \arctan 4t , \,dv = dt , v = t\,dt , \,du = \frac{4}{16t^2 + 1}
\int\arctan 4t\,dt = t\arctan 4t - \int\frac{4t}{16t^2 + 1}\,dt = t\arctan 4t - \frac{1}{8}\ln(16t^2 + 1) + C

Math Is Hard
Oct10-04, 08:13 PM
yup! looks good to me!