How do I take the indefinite integral of (5-x)/(2x^2+x-1)? I've tried splitting it into two integrals and even completing the square on the bottom but neither of those led me anywhere. Help?
Never mind! It's a partial fraction expansion problem! I got it :)
http://www.wolframalpha.com/widgets/view.jsp?id=ec4a062bb304f88c2ba0b631d7acabbc
To take the indefinite integral of (5-x)/(2x^2+x-1), you can employ the method of partial fractions.
Step 1: Factorize the denominator:
2x^2 + x - 1 = (2x - 1)(x + 1)
Step 2: Express the rational function as a sum of partial fractions:
(5 - x)/(2x^2 + x - 1) = A/(2x - 1) + B/(x + 1)
Step 3: Clear the denominators:
(5 - x) = A(x + 1) + B(2x - 1)
Step 4: Solve for A and B by equating the coefficients of like terms.
Expanding the right side and matching the coefficients on each side, we get:
5 - x = (A + 2B)x + (A - B)
Equating the coefficients of x on both sides, we have A + 2B = -1.
Equating the constants on both sides, we have A - B = 5.
Solving these equations simultaneously, we find A = 3 and B = -2.
Step 5: Rewrite the original integral using partial fractions:
∫(5 - x)/(2x^2 + x - 1) dx = ∫(3/(2x - 1) - 2/(x + 1)) dx
Step 6: Integrate each term separately:
∫(3/(2x - 1) - 2/(x + 1)) dx = 3∫(1/(2x - 1)) dx - 2∫(1/(x + 1)) dx
Step 7: Take the integral of each term using the natural logarithm property:
3∫(1/(2x - 1)) dx = 3 * (1/2) * ln|2x - 1| + C1
-2∫(1/(x + 1)) dx = -2 * ln|x + 1| + C2
where C1 and C2 are constants of integration.
Therefore, the indefinite integral of (5 - x)/(2x^2 + x - 1) is:
(3/2) * ln|2x - 1| - 2 * ln|x + 1| + C,
where C = C1 + C2 is the overall constant of integration.