Integral plz help?
whats the integral of
1/(x^2 -x) dx
2/(x^2-1) dx
please show me the steps for both, greatly appreciated!
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whats the integral of
1/(x^2 -x) dx
2/(x^2-1) dx
please show me the steps for both, greatly appreciated!
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Answers & Comments
Verified answer
You need to use partial fractions:
∫1 / (x² - x) dx
= ∫1 / x(x - 1) dx
= ∫A / x + B/(x - 1) dx
A(x - 1) + Bx = 1
(A + B)x = 0x
A = -B
-A = 1
A = -1
B = 1
= ∫-1 / x + 1/(x - 1) dx
= ∫1/(x - 1) -1 / x dx
= ln |x - 1| - ln |x| + C
= ln | (x - 1) / x | + C
∫2 / (x² - 1) dx
= ∫2 / [(x - 1)(x + 1)] dx
= ∫A / (x - 1) + B / (x + 1) dx
A(x + 1) + B(x - 1) = 2
(A + B)x = 0x
A = -B
A - B = 2
A + A = 2
2A = 2
A = 1
B = -1
= ∫1 / (x - 1) - 1 / (x + 1) dx
= ln |x - 1| - ln |x + 1| + C
= ln | (x - 1) / (x + 1) | + C