Exercise 7.4 - Question 18
Problem
$$\int \frac{5x-2}{1 + 2x + 3x^2} \, dx$$
Step-by-Step Solution
Step 1: Express Numerator
Derivative of denominator: d/dx(1 + 2x + 3x²) = 2 + 6x
Write 5x - 2 = A(6x + 2) + B
Comparing: 6A = 5, so A = 5/6; and 2A + B = -2, so B = -2 - 10/6 = -11/3
Step 2: Split the Integral
$$\int \frac{5x-2}{1+2x+3x^2} \, dx = \frac{5}{6}\int \frac{6x+2}{1+2x+3x^2}
\, dx - \frac{11}{3}\int \frac{1}{1+2x+3x^2} \, dx$$
Step 3: First Integral
Let u = 1 + 2x + 3x², du = (6x + 2)dx:
$$I_1 = \frac{5}{6}\ln|1 + 2x + 3x^2|$$
Step 4: Second Integral
Complete the square: 1 + 2x + 3x² = 3[(x + 1/3)² + 2/9]
$$I_2 = \frac{11}{3} \cdot \frac{1}{3} \cdot
\frac{3}{\sqrt{2}}\tan^{-1}\left(\frac{3x+1}{\sqrt{2}}\right) =
\frac{11}{3\sqrt{2}}\tan^{-1}\left(\frac{3x+1}{\sqrt{2}}\right)$$
✅ Final Answer
$$= \frac{5}{6}\ln|1+2x+3x^2| -
\frac{11}{3\sqrt{2}}\tan^{-1}\left(\frac{3x+1}{\sqrt{2}}\right) + C$$