∫ (arctan(x))/(1+x^2) dx?

Question

I'm very new to intergral substitution. Can someone explain what to set u equal to? I am having trouble setting this up.

Answer 1

Let u = arctan x, du= d x/(1+x²).
So the integral becomes
∫ u du = u²/2 = (arctan x)²/2 + C

Answer 2

Use regular substitution
Let u = arctan(x)
du = 1/(1+x^2) dx

Then your integral becomes

S u du = (u^2)/2 = ((arctan(x))^2)/2 + C

Answer 3

x= tanu

then arctan(x) = arctan(tan[u]) = u
and
1 + (tan [u] )^2 = (sec [u] )^2
and
dx = d(tan [u] ) = (sec [u])^2 du

∫ (arctan(x))/(1+x^2) dx
=∫ { u / (sec [u] )^2 } * (sec [u])^2 du
= (u^2) / 2
=(arctan [x] )^2 /2

Answer 4

45b