What is #int -x/ (x^2 -2x +1)dx#?

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Answer 1 · 2018-03-19T22:44:13

#int-x/(x^2-2x+1)dx=1/(x-1)+lnabs(1/(x-1))+"c"#

In order to find #int-x/(x^2-2x+1)dx#, we must perform a partial fraction decomposition on the integrand.

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

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

So

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

#=1/(x-1)-lnabs(x-1)+"c"#

#=1/(x-1)+lnabs(1/(x-1))+"c"#