Question #ff170

#int(6x-6)/(x^2+2x+2)dx#

#=int(6x-6)/((x^2+2x+1)+1)dx#

#=int(6x-6)/((x+1)^2+1)dx#

In order to make the denominator resemble the trig identity #tan^2theta+1=sec^2theta#, let #tantheta=x+1#.

This also imply that #sec^2thetacolor(white).d theta=dx# and #x=tantheta-1#. Substituting these in gives:

#=int(6(tantheta-1)-6)/(tan^2theta+1)(sec^2thetacolor(white).d theta)#

#=int(6tantheta-12)/sec^2theta(sec^2thetacolor(white).d theta)#

#=int(6tantheta-12)d theta#

Both of these are standard integrals. If you forget the integration of #tantheta#, recall that #inttanthetacolor(white).d theta=intsintheta/costhetad theta#, then use the substitution #u=costheta#.

#=-6lnabscostheta-12theta+C#

Our original was substitution was #tantheta=x+1#, so #theta=tan^-1(x+1)#.

Furthermore, if #tantheta=x+1#, this is a triangle where the side opposite #theta# is #x+1# and the side adjacent is #1#, so the hypotenuse is #sqrt((x+1)^2+1)=sqrt(x^2+2x+2)#.

Then, #costheta="adjacent"/"hypotenuse"=1/sqrt(x^2+2x+2)#. So:

#=-6lnabs(1/sqrt(x^2+2x+2))-12tan^-1(x+1)#

Using #1/sqrt(x^2+2x+2)=(x^2+2x+2)^(-1/2)# and the log rule #ln(a^b)=bln(a)#, this becomes:

#= color(blue)(3ln(x^2+2x+2)-12tan^-1(x+1)+C#

The absolute value bars aren't necessary because #x^2+2x+2>0# for all real values of #x#.