## KROD53 How do you integrate (x-1)/(x^2 -4x + 5)? one year ago one year ago

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$\int\limits_{ }^{} \frac{ x-1 }{x ^{2}-4x+5 }$

$\frac{ 1 }{2 } \int\limits_{}^{} \frac{ 2x-4+2 }{ x^{2}-4x+5 }$

$\frac{ 1 }{ 2 } \int\limits_{}^{} \left[ \int\limits_{}^{} \frac{ 2x-4 }{ x ^{2}-4x+5 } dx + \int\limits_{}^{} \frac{ 2 }{ x ^{2}-4x+5 }dx\right]$

sorry we have not integral side $\frac{ 1 }{ 2 }$

$\frac{ 1 }{ 2 } \left[ \ln (x ^{2}-4x+5) + 2\int\limits_{}^{}\frac{ 1 }{ x ^{2} -4x+5 } dx\right]$

we have $\int\limits_{}^{} \frac{ 1 }{ (x-2)^{2} +1}dx = \tan^{-1} (x-2)$

so the integral is $\frac{ 1 }{ 2 }\ln (x ^{2}-4x+5) + \tan^{-1} (x-2) +C$

$\int\limits x-1\div x ^{2}-4x + 4 +4 -8 +5$

$\int\limits \frac{ x-1 }{ (x-2)^2 +1 }$

$\int\limits \frac{ x }{ (x-2)^2 +1 }dx - \int\limits \frac{ 1 }{ (x-2)^2 +1 } dx$

$\ln (x^2 -4x +5) - \tan^{-1} (x-2) +C$