What is the antiderivative of arctan(9x)/(1+81x^2)

what are the steps?? i would know how to solve it if the problem was arctan(x)/(1+x^2) b/c i would use "u substitution" and let u=arctan(x) and du=1/(1+x^2)dx and so on... but the 9 and the 81 throw me off!

It's basically the same thing, other than the fact that the derivative of arctan(9x) is:
[(d/dx)(9x)] * 1/[1 + (9x)^2] = 9/(1 + 81x^2), by the Chain Rule.

So, if u = arctan(9x) dx, then du = 9/(1 + 81x^2) dx
==> (1/9)du = 1/(1 + 81x^2) dx.

Thus, if you substitute u = arctan(9x) and (1/9)du = 1/(1 + 81x^2) into the integral and compute the result like you would with ∫ arctan(x)/(1 + x^2) dx.

I hope this helps!

Glad I could help.

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Let u = arctan(9x)

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

du/9 = 1 / 1+81x^2

(1/9)∫ u du

(1/9)(1/2)u^2 + C

Then just back sub for u:
(1/8) (arctan(9x))^2 + C

∫arctan(9x)/(1 + 81x^2) dx

u = arctan(9x) => du = 9/(1 + 81x^2) dx

du/9 = dx/(1 + 81x^2)

1/9*∫u du => (u²)/18 + C

arctan²(9x)/18 + C