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| - ln |x+3|] + C
= 1/ 6 ln |(x-3)/(x+3)| + C..........that is ya answer
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
The answers are the same.
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
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
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Answers & Comments
Verified answer
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
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
The answers are the same.
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
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