How do you find the primitive/integral of

Tiffany Maldonado

Tiffany Maldonado

Answered question

2022-04-10

How do you find the primitive/integral of arctanx?
I tried searching for how you derive the integral/primitive of arctan(x), but I can't find any question on S.E with an answer that clearly explains this. I feel that there should be one, since it is rather fundamental knowledge, and it does come up every now and then as part of S.E questions.
So the primitive can be looked up and it should be the following:
arctan(x)dx=xarctan(x)12ln(1+x2)+C
How do you get the result?

Answer & Explanation

Quimpoah3b

Quimpoah3b

Beginner2022-04-11Added 13 answers

The integral can be solved by using integration by parts. The trick is to think of the function as a product of arctan(x) and 1, both with respect to x.
We solve the integral by choosing to integrate 1 to x and differentiate arctan(x) into 11+x2 which is easier to integrate:
arctan(x)1dx=xarctan(x)x1+x2dx+C0
and we can use the fact that 12(ddxln(1+x2))=x1+x2 to integrate x1+x2 so the last integration results in the given primitive:
xarctan(x)x1+x2dx=xarctan(x)12ln(1+x2)+C
Egerlandsq0z

Egerlandsq0z

Beginner2022-04-12Added 14 answers

Indeed, if you know a primitive for f, say F, you can derive a primitive for f1 in the same way: putting x=f(u), so u=f1(x), we have dx=f(u)du, so
f1(x)dx=uf(u)du
=uf(u)f(u)du
=f(u)uF(u)+C
=xf1(x)F(f1(x))+C
For example, in this case f(u)=tanu, so a primitive is
log|sec{u}|=12log{(1+tan2{u})}=12log{(1+x2)} , and the result follows. One can apply the same rule for logx, and all the other trigonometric and hyperbolic inverses.

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