What is the integral of #1/(1+x^2)#?

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Answer 1 · 2018-06-28T13:48:21

#int1/(1+x^2)dx=tan^-1x+C#

#color(blue)(int(du)/(1+u^2)=tan^-1u+C##rarr# Where #u# is a function of #x#

#color(red)("Proof:")#

#int(du)/(1+u^2)#

#u=tantheta##rarr##du=sec^2thetad(theta)#

#int(du)/(1+u^2)=int(sec^2thetad(theta))/(1+tan^2theta#

#color(green)(sec^2theta=1+tan^2theta#

#int(sec^2thetad(theta))/(1+tan^2theta)=int((cancel(1+tan^2theta))d(theta))/cancel(1+tan^2theta)#

#=intd(theta)=theta#

Reverse the Substitution

#u=tantheta##color(red)(rarr##theta=tan^-1u#

#therefore int(du)/(1+u^2)=tan^-1u+C#

Simply by Substituting in this relation

#int(dx)/(1+x^2)=tan^-1x+C#

Answer 2 · 2018-06-28T13:48:26

# arctanx+C#.

It is one of the Standard Integral : #int1/(1+x^2)=arctanx+C#.

Aliter :

Let, #I=int1/(1+x^2)dx#.

Subst. #x=tanu. :. dx=sec^2udu, &, u=arctanx#.

#:. I=int1/(1+tan^2u)sec^2udu#,

#=int1/sec^2usec^2udu#,

#=int1du#,

#=u#.

# rArr I=arctanx+C#.