Question

How do you integrate (x-1)/(x^2 -4x + 5)?

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

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

Answer 12

this is the right ans....when u have a quadratic eq. in denominator remember to add and subtract b^2/2a.....and then make quadratic + constant form eq.