What is the integral of arctan x?
Sorry. I lied.
Misread the question.
What is the indefinite integral of arccot(4x) ?
arccot is simply related to arctan:
arccot(x) = pi/2 - arctan(x)
You can integrate arctan using partial integration. You get x arctan(x) - integral of xdx/(x^2+1), the latter integrand contans in the denominater the derivative of the numerator (up to a factor 2), so you get a logarithmic function.
To find the integral of arctan x, we can use integration by substitution.
Let's start by using a substitution, letting u = arctan x. We can then differentiate both sides of this equation with respect to x:
du/dx = 1 / (1 + x^2)
Now, we can solve for dx:
dx = du / (1 + u^2)
Substituting dx and u into the original integral, we have:
∫(arctan x) dx = ∫u * (du / (1 + u^2))
Simplifying the integral, we get:
∫(arctan x) dx = ∫(u / (1 + u^2)) du
To evaluate this integral, we can perform another substitution. Let's use v = 1 + u^2. Differentiating both sides with respect to u, we get:
dv/du = 2u
Now, solving for u, we have:
u = (1/2) * (dv/du)
Substituting u back into the integral, we get:
∫(u / (1 + u^2)) du = ∫(((1/2) * (dv/du)) / v) du
Simplifying, we have:
∫(((1/2) * (dv/du)) / v) du = (1/2) * ∫((1 / v) dv
The integral on the right side is a straightforward integral, giving:
(1/2) * ln |v| + C
Finally, substituting back for v, we get:
(1/2) * ln |1 + u^2| + C
Replacing u with arctan x, we have the final answer:
∫(arctan x) dx = (1/2) * ln |1 + (arctan x)^2| + C