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

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Answer 1 · 2016-03-22T02:37:47

#arctan(x-1)+C#

We want to find:

#intdx/(x^2-2x+2)#

You will want to recognize that this is close to fitting the form of the #arctan# integral, which is, for future reference:

#int(du)/(u^2+a^2)=1/aarctan(u/a)+C#

In order to write the denominator of the integral #x^2-2x+2#, we must complete the square. We will want to use the perfect square #x^2-2x+1=(x-1)^2#.

#=intdx/(x^2-2x+1+2-1)#

This is equal to the original integral expression—all that's been done is #+1# and then #-1# to balance it out.

Now, group the #(x^2-2x+1)# and #(2-1)#:

#=intdx/((x^2-2x+1)+(2-1))#

#=intdx/((x-1)^2+1)#

We can now apply the #arctan# integral, if we let

#{(u=x-1" "=>" "du=dx),(a=1):}#

This gives us

#=int(du)/(u^2+a^2)#

Which equals

#=1/aarctan(u/a)+C=1/1arctan((x-1)/1)+C#

#=arctan(x-1)+C#

Answer 2 · 2017-06-22T05:23:30

We see that:

#intdx/(x^2-2x+2)=intdx/((x-1)^2+1)#

Let #x-1=tantheta#. This implies that #(x-1)^2+1=tan^2theta+1=sec^2theta# and that #dx=sec^2thetad theta#. Then the integral can be expressed as:

#=int(sec^2thetad theta)/sec^2theta=intd theta=theta+C#

From #tantheta=x-1# we see that #theta=arctan(x-1)#:

#=arctan(x-1)+C#