What is #int ( 1+x)/(1+x^2)dx#?

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Answer 1 · 2018-03-23T06:49:06

The answer is #=arctan(x)+1/2ln(1+x^2)+C#

The integral is

#int((1+x)dx)/(1+x^2)=int(1dx)/(1+x^2)+int(xdx)/(1+x^2)#

#I=I_1+I_2#

#I_1# is a standard integral

#I_1=arctan(x)#

For #I_2#, perform the substitution

#u=1+x^2#, #=>#, #du=2xdx#

Therefore,

#I_2=1/2int(du)/(u)=1/2ln(u)#

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

And finally,

#int((1+x)dx)/(1+x^2)=arctan(x)+1/2ln(1+x^2)+C#