Integrating $\int\arctan 4t \,dt$ - Integration by Parts

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To integrate ∫arctan(4t) dt, integration by parts is applied with u = arctan(4t) and dv = dt. This results in v = t and du = 4/(16t^2 + 1) dt. The integration by parts formula yields ∫arctan(4t) dt = t*arctan(4t) - ∫(4t/(16t^2 + 1)) dt. The final result is t*arctan(4t) - (1/8)ln(16t^2 + 1) + C, confirming the integration process is correct.
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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.
 
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Let u = arctan4t
dv = dt
 
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
 
yup! looks good to me!
 
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