integral of (2x) / [(x+2)(x+1)]
Can you do this with partial fractions?
thank you!!!
To find the integral of the function (2x) / [(x+2)(x+1)], we need to decompose the expression into partial fractions. The expression in the denominator can be factored as (x+2)(x+1).
To decompose the fraction, we start by writing it as the sum of two fractions:
(2x) / [(x+2)(x+1)] = A / (x+2) + B / (x+1)
Where A and B are constants that we need to determine.
Now, to find the values of A and B, we need to clear the denominators:
2x = A(x+1) + B(x+2)
Expanding the right side:
2x = Ax + A + Bx + 2B
Now, we can gather the terms with x and the constants separately:
(2x) = (A + B)x + A + 2B
This equation holds for all values of x. By comparing the coefficients of x and the constants on both sides, we can form a system of equations to solve for A and B.
Comparing the coefficients of x:
2 = A + B ...(Equation 1)
Comparing the constants:
0 = A + 2B ...(Equation 2)
Now, solving this system of equations, we can find the values of A and B:
From Equation 2, we have A = -2B.
Substituting this value back into Equation 1, we get:
2 = -2B + B
2 = -B
B = -2
Now, substituting the value of B into Equation 2, we have:
0 = A + 2*(-2)
0 = A - 4
A = 4
So, A = 4 and B = -2.
Plugging these values back into the partial fraction decomposition, we get:
(2x) / [(x+2)(x+1)] = 4 / (x+2) - 2 / (x+1)
Now, we can integrate the decomposed fractions separately:
∫(2x) / [(x+2)(x+1)] dx = ∫4 / (x+2) dx - ∫2 / (x+1) dx
Using the property of logarithmic integration, the antiderivative of 4 / (x+2) is 4ln|x+2|, and the antiderivative of 2 / (x+1) is 2ln|x+1|.
So, the integral becomes:
∫(2x) / [(x+2)(x+1)] dx = 4ln|x+2| - 2ln|x+1| + C
Where C is the constant of integration.