What is #int 2/(4+x^(2)) dx#?

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Answer 1 · 2018-04-11T06:15:22

The answer is #=arctan(x/2)+C#

Perform this integral by substitution

#4+x^2=4(1+(x/2)^2)#

Let #tanu=x/2#

#sec^2udu=1/2dx#

#1+tan^2u=sec^2u#

Therefore, the integral is

#int(2dx)/(4+x^2)=int(4sec^2udu)/(4(1+tan^2u))#

#=int(sec^2udu)/(sec^2u)#

#=int(du)#

#=u#

#=arctan(x/2)+C#

Answer 2 · 2018-04-11T06:19:51

The integral is equal to #arctan(x/2)+C#.

To solve the integral, use the substitution #u=x/2#, which means #x=2u# and #dx=2du#:

#color(white)=int2/(4+x^2)# #dx#

#=int2/(4+(2u)^2)# #2du#

#=int4/(4+4u^2)# #du#

#=intcolor(red)cancelcolor(black)4/(color(red)cancelcolor(black)4(1+u^2))# #du#

#=int1/(1+u^2)# #du#

#=arctan(u)+C#

#=arctan(x/2)+C#

That's the integral. Hope this helped!