INTEGRATION. Why both results are the same?

Question

$4 ln (x^{2} - 4) + x^{2} - 4 + c$ $4 ln(x^2-4) + x^2 - 4 + c$
and
$4 ln (x^{2} - 4) + x^{2} + c$ $4 ln(x^2-4) + x^2 + c$

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Answer 1 · 2018-06-07T16:58:17

See explanation

Remember $- 4$ $color(red)(-4$ is just a constant, so

$4 ln (∣ ∣ x^{2} - 4 ∣ ∣) + x^{2} - 4 + C$ $4ln(abs(x^2-4))+x^2-4+C$

$= 4 ln (∣ ∣ x^{2} - 4 ∣ ∣) + x^{2} + C_{1}$ $=4ln(abs(x^2-4))+x^2+C_1$

Where, $C_{1} = - 4 + C$ $color(blue)(C_1=-4+C$