evaluate integral or state that it is diverges
integral -oo, -2 [2/(x^2-1)] dx
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integral -oo, -2 [2/(x^2-1)] dx
Through partial fractions, I came up with lim [ln(x-1)-ln(x+1)] b, -2
b->-oo
I get (ln(3)-0)-(oo-oo)). The answer in the back of the book is ln(3). What am I doing wrong?
Hi there,
lim [ln(x-1)-ln(x+1)] b, -2 cannot get (ln(3)-0)-(oo-oo)) because ln(-00) has no definition. We should make it ln(x-1)-ln(x+1)=ln[(x-1)/(x+1)],then lim[ln[(x-1)/(x+1)]]b,-2 b->-00 is ln(3).
Using the difference quotient of (x+h)and Evaluate f(x)=(x+y)^3
To evaluate the integral, you correctly used partial fractions to decompose the integrand as:
2/(x^2 - 1) = (A/(x-1)) + (B/(x+1))
This gives you:
integral -oo, -2 [2/(x^2-1)] dx = integral -oo, -2 [(A/(x-1)) + (B/(x+1))] dx
Now, you can find the values of A and B by multiplying the decomposed form by the original denominator, which is (x^2 - 1):
2 = A(x + 1) + B(x - 1)
To solve for A and B, you can substitute in values of x that make one term vanish and solve for the other term. One choice is x = 1, which makes the A term vanish. Another choice is x = -1, which makes the B term vanish. So you end up with:
2 = B(-1 - 1)
2 = -2B
B = -1
Now, substitute this value of B back into the equation to solve for A:
2 = A(1 + 1)
2 = 2A
A = 1
So the decomposition becomes:
2/(x^2 - 1) = (1/(x-1)) + (-1/(x+1))
Now, you can evaluate the integral:
integral -oo, -2 [2/(x^2-1)] dx = integral -oo, -2 [(1/(x-1)) + (-1/(x+1))] dx
Remember to use the properties of logarithms when integrating. The integral of 1/(x-1) is ln|x-1|, and the integral of -1/(x+1) is -ln|x+1|. So the integral becomes:
[ln|x-1| - ln|x+1|] evaluated from -oo to -2.
Now let's evaluate the limits separately:
lim [ln|x-1| - ln|x+1|] as x approaches -oo = ln|-oo - 1| - ln|-oo + 1|
= ln|-oo| - ln|-oo|
= oo - oo
= 0
lim [ln|x-1| - ln|x+1|] as x approaches -2 = ln|-2 - 1| - ln|-2 + 1|
= ln|-3| - ln|-1|
= ln(3) - ln(1)
= ln(3) - 0
= ln(3)
Therefore, the evaluated integral is ln(3).
It seems like there was an error in your calculation. When evaluating the limit as x approaches -oo, both ln|-oo - 1| and ln|-oo + 1| are infinity, not zero. So you should have obtained oo - oo, which is undefined. This contradicts the answer in the book, which is ln(3).
Double-check your calculations to make sure there are no mistakes in finding the limits.