Question

Evaluate the integral: \[ \int \frac{2}{2 - 3x} \, dx \] The correct answer is:

• A. \( -\log |2 - 3x| + c \)
• B. \( -\frac{3}{1} \log |2 - 3x| + c \)
• C. \( -\log |2 - 3x| + c \)
• D. \( 2 \tan^{-1} \left( \frac{x}{4} \right) + c \)

Hint:

For integrals of the form \(\int \frac{1}{ax + b} \, dx\), use substitution \( u = ax + b \) and simplify the expression accordingly.

Answer 1

The Correct Option is C

We are asked to evaluate the integral: \[ I = \int \frac{2}{2 - 3x} \, dx. \] To solve this, perform the substitution: \[ u = 2 - 3x, \quad du = -3dx. \] So, \[ dx = -\frac{du}{3}. \] Substitute into the integral: \[ I = \int \frac{2}{u} \cdot \left( -\frac{du}{3} \right) = -\frac{2}{3} \int \frac{1}{u} \, du. \] Now, the integral of \(\frac{1}{u}\) is \(\ln |u|\), so we get: \[ I = -\frac{2}{3} \ln |u| + c. \] Substitute \(u = 2 - 3x\) back: \[ I = -\frac{2}{3} \ln |2 - 3x| + c. \] Thus, the correct answer is \( (C) \).