How do you find the integral #ln(x^2 + 2x + 2)dx#?

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Answer 1 · 2018-05-15T06:30:46

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

First complete the square

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

Let

#x+1=u#, #=>#, #du=dx#

Therefore,

#I=intln(x^2+2x+2)dx=intln((x+2)^2+1)dx#

#=intln(u^2+1)du#

#p=ln(u^2+1)#, #=>#, #p'=(2u)/(u^2+1)#

#q'=1#, #=>#, #q=u#

Therefore,

#I=u ln(u^2+1)-int(2u^2du)/(u^2+1)#

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

#=2intdu-2int(du)/(u^2+1)#

#=2u-2arctan(u)#

Finally,

#I=u ln(u^2+1)-2u+2arctan(u)#

#=(x+1)ln(x^2+2x+2)-2(x+1)+2arctan(x+1)+C#