Question

Evaluate \( \int \frac{2x-3}{x^2+1} dx \).

Hint:

When the degree of the numerator is one less than the denominator, try to create the derivative of the denominator in the numerator to simplify integration.

Answer 1

Step 1: Understanding the Concept:
The integral can be split into two separate parts that correspond to standard integration forms: the derivative of the denominator in the numerator, and the standard inverse trigonometric form.
Step 2: Key Formula or Approach:
1. \( \int \frac{f'(x)}{f(x)} dx = \log|f(x)| + C \).
2. \( \int \frac{1}{x^2+1} dx = \tan^{-1} x + C \).
Step 3: Detailed Explanation:
Split the integral:
\[ I = \int \frac{2x-3}{x^2+1} dx = \int \frac{2x}{x^2+1} dx - \int \frac{3}{x^2+1} dx \] For the first part \( \int \frac{2x}{x^2+1} dx \):
Let \( u = x^2+1 \implies du = 2x dx \).
\[ \int \frac{1}{u} du = \log|u| = \log(x^2+1) \] (Note: \( x^2+1 \) is always positive, so absolute value bars are not strictly necessary).
For the second part \( \int \frac{3}{x^2+1} dx \):
\[ 3 \int \frac{1}{x^2+1} dx = 3 \tan^{-1} x \] Combining the results and adding the constant of integration \( C \):
\[ I = \log(x^2+1) - 3\tan^{-1} x + C \] Step 4: Final Answer:
The evaluated integral is \( \log(x^2+1) - 3\tan^{-1} x + C \).