integral (4x^2-3x+2)/(4x^2-4x+3) dx
Why did the scarecrow become a mathematician? Because he wanted to solve complex integrals like this one!
But don't worry, I'm here to help. Let's break down this integral step by step.
To evaluate this integral:
∫ (4x^2 - 3x + 2) / (4x^2 - 4x + 3) dx,
First, let's try to simplify the expression by factoring the denominator. But oh, wait! The denominator doesn't easily factorize. Looks like we need another approach.
What you can do is use a technique called partial fraction decomposition. This involves splitting the fraction into multiple smaller fractions. Sounds complicated, right? Don't worry, I'll 'simplify' it for you.
Let's say the original fraction can be expressed in the following way:
(4x^2 - 3x + 2) / (4x^2 - 4x + 3) = A / (Bx - C) + D / (Ex - F),
Now, to find the values of A, B, C, D, E, and F, you need to use algebraic manipulation. Unfortunately, I'm more skilled in juggling than in solving simultaneous equations, so I'll have to leave that part up to you.
Once you find the values of A, B, C, D, E, and F, the integral can be split into individual fractions, like this:
∫ (4x^2 - 3x + 2) / (4x^2 - 4x + 3) dx = A ∫ dx / (Bx - C) + D ∫ dx / (Ex - F).
These individual fractions are much easier to integrate because they are of the form ∫ dx / x, which evaluates to natural log |x|.
So, after calculating A, B, C, D, E, and F, you can substitute them back into the integral expression and evaluate each term separately.
But remember, this process might involve some complex calculations, so don't let the numbers clown around with you!
To solve the integral:
∫ (4x^2 - 3x + 2)/(4x^2 - 4x + 3) dx
Step 1: Check if the denominator factors.
The denominator does not factor easily, so we need to use a different method.
Step 2: Complete the square for the denominator.
Let's rewrite the denominator in the form of (a + b)^2 + c:
4x^2 - 4x + 3 = (2x - 1)^2 + 2
Step 3: Rewrite the integral using the substitution.
We'll let u = 2x - 1, which means du = 2dx.
Step 4: Substitute the new variable into the integral.
∫ (4x^2 - 3x + 2)/(4x^2 - 4x + 3) dx
= ∫ (1/2) (4(2x - 1)^2 + 2)/(4x^2 - 4x + 3) dx
= (1/2) ∫ (4u^2 + 2)/(u^2 + 2) du
Step 5: Simplify the integral.
Now, we have a simpler integral to solve.
∫ (4u^2 + 2)/(u^2 + 2) du = ∫ (4u^2/(u^2 + 2) + 2/(u^2 + 2)) du
Step 6: Split the integral into two separate integrals.
∫ (4u^2/(u^2 + 2)) du + ∫ (2/(u^2 + 2)) du
Step 7: Solve each integral separately.
For the first integral:
∫ (4u^2/(u^2 + 2)) du
Let's make a substitution v = u^2 + 2. This means dv = 2u du.
∫ (2u dv)/v = 2 ∫ (u/v) dv
= 2 ln|v| + C
= 2 ln|u^2 + 2| + C
For the second integral:
∫ (2/(u^2 + 2)) du
Let's make a substitution w = u/√2. This means du = √2 dw.
∫ (2/((√2)^2 * (w^2 + 1))) du
= ∫ (2/2(w^2 + 1)) du
= ∫ (1/(w^2 + 1)) du
= arctan(w) + C
= arctan(u/√2) + C
Step 8: Combine the results.
The final result is:
(1/2)(2 ln|u^2 + 2| + 2 arctan(u/√2)) + C
Step 9: Substitute the original variable back.
Since u = 2x - 1, the final answer is:
(1/2)(2 ln|(2x - 1)^2 + 2| + 2 arctan((2x - 1)/√2)) + C
To evaluate the integral of a rational function like this, you can use partial fraction decomposition. The numerator of the rational function is a polynomial of degree 2, while the denominator is also a polynomial of degree 2. Before we can perform partial fraction decomposition, we need to check if the denominator polynomial can be factored into linear factors.
The denominator is given as 4x^2 - 4x + 3. To factor this polynomial, we can use the quadratic formula.
The quadratic formula gives us x = (-b ± √(b^2 - 4ac)) / 2a, where the quadratic equation is ax^2 + bx + c = 0.
In this case, a = 4, b = -4, and c = 3. Plugging these values into the quadratic formula, we get:
x = (4 ± √((-4)^2 - 4(4)(3))) / (2(4))
= (4 ± √(16 - 48)) / 8
= (4 ± √(-32)) / 8
= (4 ± 4i√2) / 8
= (1 ± i√2) / 2
Since the discriminant is negative, the polynomial cannot be factored into linear factors. Therefore, we cannot use partial fraction decomposition.
In this case, we need to resort to a different approach. We can try using a trigonometric substitution.
Let's substitute x = tan(theta). This implies dx = sec^2(theta) d(theta). We can rewrite the integral in terms of theta.
The integral becomes:
∫ [(4tan^2(theta) - 3tan(theta) + 2) / (4tan^2(theta) - 4tan(theta) + 3)] * sec^2(theta) d(theta)
Now, we can simplify this expression.
long division shows that we have
∫ 1 + (x-1)/(4x^2-4x+3)
= ∫ 1 + (x-1)/[(2x-1)^2 + 2] dx
let (2x-1)^2 = 2tan^2θ, so
(2x-1)^2 + 2 = 2tan^2θ+2 = 2sec^2θ
2x-1 = √2 tanθ
x = (√2 tanθ+1)/2
x-1 = (√2 tanθ-1)/2
dx = 1/2 √2 sec^2θ
∫ 1 + (x-1)/[(2x-1)^2 + 2] dx
= ∫ (1 + (√2 tanθ-1)/(4sec^2θ)) 1/√2 sec^2θ dθ
. . .
= 1/8(x + 1/8 log(4x^2-4x+3) - 1/(4√22) arctan (2x-1)/√2