what's the integral of dx/(x-2)(x-2+sqrt(x-2))?

Answer 1

integral [ dx/ [ (x-2)((x -2) + sqrt(x-2 ) ] ] =

[ dx / [ (x-2) sqrt(x-2) ( sqrt(x-2) + 1) ]]

let u = sqrt(x-2)

then

du = dx/(2sqrt(x-2))

2 du = 1/sqrt(x-2)

and

u^2 = x-2

integral becomes

2 [ du/ u^2( u+1) ]

partial fraction decomposition

(Au+b)/u^2 + C/u+1 = 1/[ u^2(u+1)]

Multiplies by u^2(u+1)

(Au+b)(u+1) + c U^2 = 1

(A+c)u^2 + (A+B)u + b = 1

A+C=0

A+B=0

B = 1

then a=-1 and C = 1

(-u+1)/u^2 + 1/(u+1)

checks the numerator:

(1-u)(u+1) + u^2 = 1-u^2 + u^2 = 1

yes, partial fraction decomposition is correct

integrates (1-u)/u^2 + 1/(u+1)

= 1/u^2 - 1/u + 1/(u+1)

u^(-2) - u^(-1) + (u+1)^(-1)

(-u^(-1)) - ln u + ln u+1

-1/u - ln u + ln u+1

-1/sqrt(x-2) - 1/2 ln (x-2) + ln ( sqrt(x-2) + 1)

Then multiplies by 2 and re-arranges:

2 [ ln ( sqrt(x-2) + 1) - ln(x-2)/2 - 1/sqrt(x-2) ] + c

is the anti-derivative

Checks by diff:

1/2 cancels the 2 in each derivative

1/[sqrt(x-2)[sqrt(x-2)+1]] - 1/(x-2) + 1/(x-2)^(3/2)

1/ sqrt(x-2)^(3/2) [ sqrt(x-2) + 1] =

1/ sqrt(x-2) [ (x-2) + sqrt(x-2) ]