So I'm trying to integrate a function using partial fractions. Here is the integral of interest: ∫(3x^2+5x+3)/[(x+2)(x^2+1)]dx.
Since the numerator's degree of the polynomial is lesser than that of the denominator's degree, it is clear to separate. However, once I separate and multiply like terms, I encounter obstacle when determining the new scalar numerators.
(1) ∫(3x^2+5x+3)/[(x+2)(x^2+1)]dx = ∫[A/(x+2)]+[Bx+C/(x^2+1)]dx.
(2) Multiply for common denominator to obtain numerator corresponding to (A+B)x^2+(2B+C)x+(A+2C). Solving A, B, and C for common roots from the initial polynomial, I set up (A+B) = 3, (2B+C) = 5, (A+2C)=3. My new terms are now: A=1; B=2; C=1.
(3) ∫dx/(x+2) +∫(2x+1)/(x^2+1)dx is now the new, separated function. ∫dx/(x+2) will integrate into ln|x+2|+C. Further separation is needed for the second integral, ∫(2x+1)/(x^2+1)dx = ∫2x/(x^2+1)dx +∫dx/(x^2+1). ∫dx/(x^2+1) integrates nicely into arctanx. ∫2x/(x^2+1)dx needs substitution however, which eliminates the 2x in the numerator without adding any scalars: thus ∫2x/(x^2+1)dx = ln|x^2+1|.
∫(3x^2+5x+3)/[(x+2)(x^2+1)]dx = ln|(x+2)(x^2+1)|+arctanx+C !
Looks good to me.
Your solution is almost correct. However, there is a small mistake in step (2) when you solve for A, B, and C. Let's go through the steps again and correct the mistake:
Step (1): Separate the function using partial fractions:
∫(3x^2+5x+3)/[(x+2)(x^2+1)]dx = ∫[A/(x+2)]+[Bx/(x^2+1)]+[C/(x^2+1)]dx.
Step (2): Multiply for a common denominator and equate the numerators:
The common denominator is (x+2)(x^2+1). The numerator becomes A(x^2+1) + Bx(x+2) + C(x+2).
Expanding and collecting like terms, we have (A+B)x^2 + 2Bx + (A+C)x + A+2C.
Equating the coefficients of like powers of x on both sides, we get:
A+B = 3, 2B = 5, A+C = 0, A+2C = 3.
From the equations, we can solve for A, B, and C. Solving the equations, we find A = 1, B = 5/2, C = -1/2.
Step (3): Rewrite the integral using the partial fraction decomposition:
∫(3x^2+5x+3)/[(x+2)(x^2+1)]dx = ∫[1/(x+2)] + ∫[(5/2)x/(x^2+1)] + ∫[(-1/2)/(x^2+1)]dx.
Step (4): Integrate each term separately:
∫[1/(x+2)] dx = ln|x+2| + C1.
∫[(5/2)x/(x^2+1)] dx requires a substitution. Let u = x^2 + 1, du = 2x dx.
The integral becomes ∫(5/2)(1/2) du/u = (5/4)ln|u| + C2 = (5/4)ln|x^2+1| + C2.
∫[(-1/2)/(x^2+1)] dx = (-1/2)arctan(x) + C3.
Step (5): Combine the results:
∫(3x^2+5x+3)/[(x+2)(x^2+1)]dx = ln|x+2| + (5/4)ln|x^2+1| - (1/2)arctan(x) + C.
So, the final solution is ln|x+2| + (5/4)ln|x^2+1| - (1/2)arctan(x) + C.