What is the best method to evaluate the integral of 1/(4x^2-9) dx
a) Integration by parts
b) Rewrite the integral using long division
c) Rewrite the integral using partial fractions
d) Use a substitution -----> my answer
Can you check for me, please
you can use a trig substitution, or partial fractions.
The trig substitution will give you a result in terms of inverse hyperbolic functions
The partial fractions will give a result using logs.
Of course, the hyperbolic functions involve exponentials, so the inverse functions involve logs.Not sure which is "better."
To evaluate the integral of 1/(4x^2-9) dx, you can indeed use a substitution method. Let me explain how to do it:
1. Start by identifying a suitable substitution. In this case, we can let u = 2x. Differentiate both sides to find du = 2dx.
2. Now, rewrite the integral in terms of the substitution. Substituting u and dx in terms of u into the integral, we get:
∫(1/(4x^2-9)) dx = (1/2) ∫(1/((2x)^2-3^2)) dx = (1/2) ∫(1/(u^2-9)) du.
3. Next, simplify the integrand. We recognize that the denominator is a difference of squares, so we can rewrite it as (u-3)(u+3). Now, we have:
(1/2) ∫(1/((u-3)(u+3))) du.
4. Now, we can rewrite the integrand using partial fractions. Set up the equation A/(u-3) + B/(u+3) = 1/((u-3)(u+3)), where A and B are constants. By multiplying both sides by (u-3)(u+3), we get:
A(u+3) + B(u-3) = 1.
5. Solving for A and B, we have:
2A + (-6B) = 1 (setting u = 3)
6A + (2B) = 1 (setting u = -3).
Solving the system of equations, we find A = 1/6 and B = -1/6.
6. Now, we can rewrite our integrand using partial fractions:
(1/2) ∫(1/(u-3)(u+3)) du = (1/2) * (1/6) ∫(1/(u-3)) du - (1/2) * (1/6) ∫(1/(u+3)) du.
= (1/12) ln|u-3| - (1/12) ln|u+3| + C,
where C is the constant of integration.
7. Finally, substitute back u = 2x to obtain:
(1/12) ln|2x-3| - (1/12) ln|2x+3| + C.
Therefore, the correct answer is option d) Use a substitution.