Answer
$$\frac{3}{4}\ln \left| x \right| - \tan x + C$$
Work Step by Step
$$\eqalign{
& \int {\left[ {\frac{3}{{4x}} - {{\sec }^2}x} \right]} dx \cr
& {\text{sum rule}} \cr
& = \int {\frac{3}{{4x}}} dx - \int {{{\sec }^2}x} dx \cr
& {\text{multiple constant rule}} \cr
& = \frac{3}{4}\int {\frac{1}{x}} dx - \int {{{\sec }^2}x} dx \cr
& {\text{integrate by using the rules}} \cr
& \int {\frac{1}{x}dx} = \ln \left| x \right| + C{\text{ and }}\int {{{\sec }^2}x} dx = \tan x + C \cr
& = \frac{3}{4}\ln \left| x \right| - \tan x + C \cr} $$
