(a) Express \(f(x)\) in partial fractions:
Assume \(f(x) = \frac{A}{3x+2} + \frac{Bx+C}{x^2+4}\).
Multiply through by \((3x+2)(x^2+4)\) to clear the denominators:
\(7x + 18 = A(x^2 + 4) + (Bx + C)(3x + 2)\).
Expand and collect like terms:
\(7x + 18 = Ax^2 + 4A + 3Bx^2 + 2Bx + 3Cx + 2C\).
Combine terms: \((A + 3B)x^2 + (2B + 3C)x + (4A + 2C)\).
Equate coefficients with \(7x + 18\):
\(A + 3B = 0\)
\(2B + 3C = 7\)
\(4A + 2C = 18\)
Solve the system of equations:
From \(A + 3B = 0\), \(A = -3B\).
Substitute into \(4A + 2C = 18\):
\(4(-3B) + 2C = 18\)
\(-12B + 2C = 18\)
\(2C = 12B + 18\)
\(C = 6B + 9\)
Substitute \(C = 6B + 9\) into \(2B + 3C = 7\):
\(2B + 3(6B + 9) = 7\)
\(2B + 18B + 27 = 7\)
\(20B = -20\)
\(B = -1\)
\(A = -3(-1) = 3\)
\(C = 6(-1) + 9 = 3\)
Thus, \(f(x) = \frac{3}{3x+2} + \frac{-x+3}{x^2+4}\).
(b) Find \(\int_0^2 f(x) \, dx\):
\(\int_0^2 \left( \frac{3}{3x+2} + \frac{-x+3}{x^2+4} \right) \, dx\)
\(= \int_0^2 \frac{3}{3x+2} \, dx + \int_0^2 \frac{-x+3}{x^2+4} \, dx\)
First integral: \(\int \frac{3}{3x+2} \, dx = \ln|3x+2|\) evaluated from 0 to 2:
\(\left[ \ln|3(2)+2| - \ln|3(0)+2| \right] = \ln 8 - \ln 2 = \ln 4\)
Second integral: \(\int \frac{-x+3}{x^2+4} \, dx = -\frac{1}{2} \ln(x^2+4) + \frac{3}{2} \arctan\left(\frac{x}{2}\right)\) evaluated from 0 to 2:
\(\left[ -\frac{1}{2} \ln(4+4) + \frac{3}{2} \arctan(1) \right] - \left[ -\frac{1}{2} \ln(4) + \frac{3}{2} \arctan(0) \right]\)
\(= -\frac{1}{2} \ln 8 + \frac{3}{2} \frac{\pi}{4} + \frac{1}{2} \ln 4\)
\(= -\frac{1}{2} \ln 8 + \frac{3}{8} \pi + \frac{1}{2} \ln 4\)
Combine results:
\(\ln 4 - \frac{1}{2} \ln 8 + \frac{3}{8} \pi + \frac{1}{2} \ln 4\)
\(= \frac{3}{2} \ln 2 + \frac{3}{8} \pi\)
