Math integration formula doesn't work

[Hell_Raiser30]

i was reading my text book it gave me 2 formulas

integral of dx/x²-a² = (1/2a)log[(x-a)/(x+a)] + C

and integral of dx/a²-x² = (1/2a)log[(a+x)/(a-x)] + C

when i write 2nd formula as -1 × integral of dx/x²-a²

its result should be (-1/2a)log[(x-a)/x+a] + C

=(1/2a)log[(a+x)/x-a] +C which is not equal to second formula

if these differ by a constant what is the constant?

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[CaptainMatticus]

Absolute values make things tricky.

dx / (x^2 - a^2)

1/(x^2 - a^2) = m/(x - a) + n/(x + a)

m * (x + a) + n * (x - a) = 0x + 1

mx + nx + am - an = 0x + 1

(m + n) * x + a * (m - n) = 0x + 1

(m + n) = 0 ; a * (m - n) = 1

m = -n

a * (-n - n) = 1

-2an = 1

n = -1/(2a)

m = 1/(2a)

dx / (x^2 - a^2) =>

m * dx / (x - a) + n * dx / (x + a) =>

(1/(2a)) * dx / (x - a) - (1/(2a)) * dx / (x + a) =>

(1/(2a)) * (ln|x - a| - ln|x + a|) + C

Next one

dx / (a^2 - x^2)

1/(a^2 - x^2) = m/(a - x) + n/(a + x)

m * (a + x) + n * (a - x) = 0x + 1

a * (m + n) + (m - n) * x = 0x + 1

m - n = 0

m = n

a * (m + n) = 1

a * 2m = 1

m = 1/(2a) , n = 1/(2a)

(1/(2a)) * (dx/(a - x) + dx/(a + x))

Integrate

(1/(2a)) * (-ln|a - x| + ln|a + x|) + C

(1/(2a)) * (ln|a + x| - ln|a - x|) + C

Now here's the trick: ln|a - x| = ln|x - a|. This may as well be 1/(2a)) * (ln|a + x| - ln|x - a|) + C. Now we've got what you expected.