How do calculate that? pls help me

#int(x^2+2x+3)/(x^2+2x+1) dx#

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

as #x^2+2x+1=(x+1)^2# this becomes

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

= #intdx+2int1/(x+1)^2dx#

= #x+2int1/(x+1)^2# - now let #u=x+1# then #du=dx#

and our integral becomes

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

= #x-2/u+c#

= #x-2/(x+1)+c#

First off, split the #3# into #2+1#:

#\int(x^2+2x+3)/(x^2+2x+1)dx=\int (x^2+2x+1+2)/(x^2+2x+1)dx#

Then, use the linearity of this function to get rid of the #2# and simply drop the #1# because it's a constant:

#\int(2/(x^2+2x+1)+1)dx=2\int1/(x^2+2x+1)dx+\int 1dx#

Then, try to factor out the denominator:

#2\int1/(x^2+2x+1)dx=2\int1/(x+1)^2dx+\int 1dx#

Get rid of #\int 1dx#, using the rule #\int Kdx=Kx+C#, where #K# is any constant:

#2\int1/(x+1)^2dx+\int 1dx=2\int1/(x+1)^2dx+x#

Then, let #k=(x+1)# and apply the power rule:

#2\int1/(x+1)^2dx+x=2\int1/k^2dk+x=2\int k^{-2}dk+x#

#2\int k^{-2}dk+x=2\frac{k^{-2+1}}{-2+1}+x=2\frac{k^{-1}}{-1}+x#

#2\frac{k^{-1}}{-1}+x=\frac{-2}{k}+x=\frac{-2}{x+1}+x#

And we're done! Don't forget to add a constant, though :) Hence, we can conclude that:

#\int(x^2+2x+3)/(x^2+2x+1)dx=x-\frac{2}{x+1}+C#