Arisa asked in Science & MathematicsMathematics · 9 years ago

# how to solve this question ∫dx/x2-9 ?

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• Anonymous
9 years ago

Using partial fractions:

1/(x²-9)=A/(x+3) + B/(x-3).....(i)

1 = A(x-3) + B(x+3)

So, you get A=1/6 and B= -1/6

Now:substituting back A and B in (i)

1/(x²-9) = -1/ 6(x+3) + 1/ 6(x-3)

∫dx/x2-9

=∫ -1/ 6(x+3) + 1/ 6(x-3) dx

=∫ -1/ 6(x+3) dx + ∫ 1/ 6(x-3) dx

= -1/ 6 ∫ 1/(x+3) dx + 1/ 6 ∫ 1/(x-3) dx

= -1/ 6 ln|x+3| + 1/ 6 ln |x-3| + C

= -1/ 6 ln|x+3| + 1/ 6 ln |x-3| + C

= 1/ 6 [ ln|x-3| - ln |x+3|] + C

= 1/ 6 ln |(x-3)/(x+3)| + C..........that is ya answer

• 9 years ago

Is that ∫ dx/(x² - 9) ?

If so, then

∫ dx/(x² - 9) = (-1/3) arctanh(x/3) + C.

Alternatively, you can use a partial fraction decomposition

1/(x² - 9) = (1/6)/(x - 3) - (1/6)/(x + 3).

So

∫ dx/(x² - 9) = (1/6) ln|(x-3)/(x+3)| + C

• Como
Lv 7
9 years ago

Presentation looks dodgy.

Will take a guess that is meant to read as :-

I = ∫ dx / (x² - 9) = ∫ dx / [ (x - 3) (x + 3) ]

1 / [ (x - 3)(x + 3) ] = A / (x - 3) + B / (x + 3)

1 = A (x + 3 ) + B (x - 3)

A = 1/6

B = -1/6

I = (1/6) ∫ dx / (x - 3) - (1/6) ∫ dx / (x + 3)

I = (1/6) log (x - 3) - (1/6) log (x + 3) + C

• Mark
Lv 7
9 years ago

I guess you mean ∫ dx/(x²–9)

Partial fractions is the most direct way. (You can also use inverse hyperbolic functions if you know the formulas, but this is the more elementary method.)

1/(x²–9) = 1/(x+3)(x–3) = (–1/6)/(x+3) + (1/6)/(x–3)

So we have (1/6) [ ∫ dx/(x–3) – ∫ dx/(x+3) ] =

(1/6) [ ln |x–3| – ln |x + 3| ] + constant