What is the antiderivative of #1/(x^2 - 2x +2)#?
2 Answers
Explanation:
We want to find:
#intdx/(x^2-2x+2)#
You will want to recognize that this is close to fitting the form of the
#int(du)/(u^2+a^2)=1/aarctan(u/a)+C#
In order to write the denominator of the integral
#=intdx/(x^2-2x+1+2-1)#
This is equal to the original integral expression—all that's been done is
Now, group the
#=intdx/((x^2-2x+1)+(2-1))#
#=intdx/((x-1)^2+1)#
We can now apply the
#{(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#
We see that:
Let
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