integral(8&4) y dy/(y^2-2y-3)
int y dy/(y^2-2y-3) =
int 1/2 2y dy/(y^2-2y-3) =
int 1/2 (2y - 2 + 2) dy/(y^2-2y-3) =
1/2 Log(y^2 - 2 y - 3) +
int dy/(y^2-2y-3)
int dy/(y^2-2y-3) =
int dy/[(y-3)(y+1)]
1/[(y-3)(y+1)] =
1/4 [1/(y-3) - 1/(y+1)] ----->
int dy/(y^2-2y-3) =
1/4 Log[(y-3)/(y+1)]
To solve the integral ∫(8 to 4) y dy / (y^2 - 2y - 3), follow these steps:
1. Factorize the denominator. The denominator y^2 - 2y - 3 can be factored as (y - 3)(y + 1).
2. Rewrite the integral as ∫(8 to 4) y dy / ((y - 3)(y + 1)).
3. Decompose the fraction into partial fractions. To do so, set it up as A / (y - 3) + B / (y + 1).
4. Cross multiplying the denominators and combining like terms, we get y = A(y + 1) + B(y - 3).
5. Expanding the equation, we have y = Ay + A + By - 3B.
6. Equating the coefficients of like powers of y, we get A + B = 1 and A - 3B = 0.
7. Solving the system of equations, we find A = 3/4 and B = 1/4.
8. Rewriting the integral with the partial fractions, we have ∫(8 to 4) (3/4)/(y - 3) + (1/4)/(y + 1) dy.
9. Integrating each term separately, we have ∫(8 to 4) (3/4) ln|y - 3| + (1/4) ln|y + 1| dy.
10. Evaluating the integral by substituting the limits, we have [(3/4) ln|y - 3| + (1/4) ln|y + 1|] from 8 to 4.
11. Substituting the limits and simplifying, we get [(3/4) ln(1) + (1/4) ln(5)] - [(3/4) ln(5) + (1/4) ln(3)].
12. Simplifying further, we have (1/4)[ln(1) - ln(3) + 3 ln(5) - 3 ln(5)].
13. Since ln(1) equals 0, the result simplifies to (1/4)(3 ln(5) - ln(3)).
Therefore, the value of the integral ∫(8 to 4) y dy / (y^2 - 2y - 3) is (1/4)(3 ln(5) - ln(3)).
To find the integral of the function f(y) = y / (y^2 - 2y - 3) over the interval [8, 4], we can follow these steps:
Step 1: Factorize the denominator:
The denominator of the function y^2 - 2y - 3 can be factored as (y - 3)(y + 1).
Step 2: Write the integral in partial fraction form:
A/(y - 3) + B/(y + 1), where A and B are constants.
Step 3: Find the values of A and B:
Multiply both sides of the equation by the original denominator (y^2 - 2y - 3) to eliminate denominators:
A(y + 1) + B(y - 3) = y
Simplify the equation and equate the coefficients of like powers of y:
(A + B)y + (A - 3B) = y
Comparing coefficients, we get two equations:
A + B = 1 (equation 1)
A - 3B = 0 (equation 2)
Solving equations 1 and 2 simultaneously, we find A = 3/4 and B = 1/4.
Step 4: Write the integral in partial fraction form with the determined values of A and B:
(3/4)/(y - 3) + (1/4)/(y + 1)
Now, we can integrate each term separately using the power rule of integration.
Integral of (3/4)/(y - 3) = (3/4) * ln|y - 3|
Integral of (1/4)/(y + 1) = (1/4) * ln|y + 1|
Step 5: Evaluate the integral using the limits of integration:
Substitute the upper and lower limits into the integrated expressions:
[(3/4) * ln|y - 3|] from 8 to 4 + [(1/4) * ln|y + 1|] from 8 to 4
Evaluate each term separately:
[(3/4) * ln|4 - 3|] + [(1/4) * ln|4 + 1|] - [(3/4) * ln|8 - 3|] - [(1/4) * ln|8 + 1|]
Simplifying, we get:
(3/4) * ln|1| + (1/4) * ln|5| - (3/4) * ln|5| - (1/4) * ln|9|
Since ln|1| = 0, we can eliminate the first term:
(1/4) * ln|5| - (3/4) * ln|5| - (1/4) * ln|9|
Further simplification yields:
(-2/4) * ln|5| - (1/4) * ln|9|
Which can be simplified to:
- (1/2) * ln|5| - (1/4) * ln|9|
Therefore, the integral of y / (y^2 - 2y - 3) from 8 to 4 is - (1/2) * ln|5| - (1/4) * ln|9|.