evaluate the integral
integral of 3 to 2 x/(x^2-2)^2 dx
u=x^2-2
du=2x dx
1/2 du = x dx
integral of 1/u^2 du
-1/(x^2-2) Then I plug in 3 and 2 and subtract them form each other
-1/(3^2-2) - (-1/(2^2-2) Is this correct?
The (1/2) has been missed out, should read:
∫ du/(2u^2)
=-1/(2(x^2-2))
After that you can plug in the limits of integration. I get 5/28 integrating from 2 to 3.
The bottom limit is the start value, and the top value is the end value.
It is usual to go from 2 to 3, and rarely from 3 to 2. Please check.
Yes, your solution is almost correct. Let's go through the steps together to ensure accuracy.
To evaluate the integral ∫(3 to 2) x/(x^2-2)^2 dx, you made a good substitution:
u = x^2 - 2.
Now, let's calculate du using the derivative of u with respect to x:
du/dx = 2x.
Rearrange the equation to solve for dx:
dx = du/(2x).
Substituting x dx with du/(2x) in the integral, we get:
∫(3 to 2) (x/(x^2-2)^2) dx = ∫(3 to 2) (1/(2x))(x/u^2) du.
Now, simplify the integral by canceling out 1/(2x):
∫(3 to 2) (1/u^2) du.
Integrating (1/u^2) with respect to u, we get:
∫(1/u^2) du = -1/u.
Now, we can evaluate the integral from 3 to 2 as follows:
-1/(3^2-2) - (-1/(2^2-2))
= -1/(9-2) - (-1/(4-2))
= -1/7 - (-1/2)
= -1/7 + 1/2
= (-2 + 7)/(7 * 2)
= 5/14.
Therefore, the value of the integral ∫(3 to 2) x/(x^2-2)^2 dx is 5/14.