Final answer:
To evaluate the integral ∫ arctan(9t) dt, we can use integration by substitution and integration by parts. The final result is 1/9 (u * arctan(u) - 1/2 ln|1 + u^2|) + C.
Step-by-step explanation:
To evaluate the integral ∫ arctan(9t) dt, we can use integration by substitution. Let u = 9t, then du/dt = 9 and dt = du/9. Substituting into the integral, we have ∫ arctan(u) (du/9) = 1/9 ∫ arctan(u) du.
Now we can integrate ∫ arctan(u) du. Let's use integration by parts with u = arctan(u) and dv = du. Then du = 1/(1 + u^2) du and v = ∫dv = u.
Applying the integration by parts formula, we have ∫ arctan(u) du = u * arctan(u) - ∫ u(1/(1 + u^2)) du. The second integral can be simplified using a trigonometric substitution. Let 1 + u^2 = v, then 2u du = dv. Substituting, we have ∫ u(1/(1 + u^2)) du = 1/2 ∫ (1/v) dv = 1/2 ln|v| = 1/2 ln|1 + u^2|.
Substituting back in for u, we have 1/9 ∫ arctan(u) du = 1/9 (u * arctan(u) - 1/2 ln|1 + u^2|) + C.