How do you integrate $int (2x-2)/(x^2 - 6x +10)^(1/2)dx$ using partial fractions?

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Answer 1 · 2015-11-22T15:16:28

$[2sqrt(x^2-6x+10)]+[4arcsinh(x-3)]+C$

$int (2x-2)/sqrt(x^2 - 6x +10)dx$

$int (2x-6+4)/sqrt(x^2 - 6x +10)dx$

$int (2x-6)/sqrt(x^2 - 6x +10)dx+4int1/sqrt(x^2-6x+10)dx$

For the first integral let's $u = x^2-6x+10$

$du = 2x-6 dx$

we have $int1/sqrt(u)du$

which is $[2sqrt(u)]$

complete the square for the second

$x^2-6x+10 = x^2-6x+10 + 9 - 9 = (x-3)^2+1$

$[2sqrt(x^2-6x+10)]+4int1/sqrt((x-3)^2+1)dx$

let's $t = x-3$

$dt = dx$

$[2sqrt(x^2-6x+10)]+4int1/sqrt(t^2+1)dx$

$[2sqrt(x^2-6x+10)]+[4arcsinh(x-3)]+C$

How do you integrate int (2x-2)/(x^2 - 6x +10)^(1/2)dxint (2x-2)/(x^2 - 6x +10)^(1/2)dx using partial fractions?