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Can someone plz show me how to show that this problem below is CORRECT? Thanks a lot.
Integral of dx/(1-x^2) = x/(1-x)
StephenPrivitera
Oct9-03, 10:04 PM
Use the substitution x=sinu.
Hello, gigi9!
The answer is not correct . . .
Originally posted by gigi9
Can someone plz show me how to show that this problem below is CORRECT?
Integral of dx/(1-x^2) = x/(1-x)
We can use Partial Fractions: 1/(1 - x^2) = A/(1 - x) + B/(1 + x)
and find that: A = 1/2, B = -1/2.
The answer will be: (1/2) ln|(1 - x)/(1 + x)| + C
Can someone plz show me how to show that this problem below is CORRECT? Thanks a lot.
Integral of dx/(1-x^2) = x/(1-x)
The easiest way to check if an integral is correct is to invert the operation and differentiate.
Via the fundamental theorem of calculus, if
∫ dx / (1 - x^2) = x / (1 - x)
then
1 / (1 - x^2) = (d/dx) (x / (1 - x))
So if we actually perform the differentiation, we get:
1 / (1 - x^2) = 1 / (1 - x)^2
Because this equation is false, the original problem (as you've written it) must be false as well.
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