- #1

Dell

- 590

- 0

^{2}-2x+2)dx

(x-5)/(x

^{2}-2x+2)=(x-1-4)/((x-1)

^{2}+1)

x-1=t therefore x=t+1

dx=x'dt=(t+1)'dt=dt

[tex]\int[/tex](x-5)/(x

^{2}-2x+2)dx=[tex]\int[/tex](t-4)/(t

^{2}+1)dt

=[tex]\int[/tex]t/(t

^{2}+1)dt-4[tex]\int[/tex]1/(t

^{2}+1)dt

=0.5ln|t

^{2}+1|-4arctg(t)+c