intregetion dx/x^2-9
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I = dx/( x² -9)
=dx/(x -3)(x +3)
=1/6{dx/(x - 3) - dx/(x + 3)}
we know , from basic formula,
dx/x = lnx + C
use this here,
= 1/6 { ln(x -3) - ln(x + 3) }
= 1/6 ln( x -3)/(x +3)
hence, integration of 1/(x² - 9)
is 1/6ln(x -3)/(x +3)
=dx/(x -3)(x +3)
=1/6{dx/(x - 3) - dx/(x + 3)}
we know , from basic formula,
dx/x = lnx + C
use this here,
= 1/6 { ln(x -3) - ln(x + 3) }
= 1/6 ln( x -3)/(x +3)
hence, integration of 1/(x² - 9)
is 1/6ln(x -3)/(x +3)
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