- #1
Dell
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- 0
[tex]\int[/tex](x-5)/(x2-2x+2)dx
(x-5)/(x2-2x+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)/(x2-2x+2)dx=[tex]\int[/tex](t-4)/(t2+1)dt
=[tex]\int[/tex]t/(t2+1)dt-4[tex]\int[/tex]1/(t2+1)dt
=0.5ln|t2+1|-4arctg(t)+c
(x-5)/(x2-2x+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)/(x2-2x+2)dx=[tex]\int[/tex](t-4)/(t2+1)dt
=[tex]\int[/tex]t/(t2+1)dt-4[tex]\int[/tex]1/(t2+1)dt
=0.5ln|t2+1|-4arctg(t)+c