Question

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

Hint:

To integrate functions of the form \( \frac{2x}{x^2 + 1} \), use substitution, and for \( \frac{1}{x^2 + 1} \), recall that it integrates to \( \tan^{-1}(x) \).

Answer 1

Step 1: Breaking the integral into two parts.
We can break the integral into two parts: \[ \int \frac{2x - 3}{x^2 + 1} \, dx = \int \frac{2x}{x^2 + 1} \, dx - \int \frac{3}{x^2 + 1} \, dx \] Step 2: Solving the first part.
The first integral is: \[ \int \frac{2x}{x^2 + 1} \, dx \] This is a standard integral. Let \( u = x^2 + 1 \), so \( du = 2x \, dx \). The integral becomes: \[ \int \frac{du}{u} = \ln|u| = \ln(x^2 + 1) \] Step 3: Solving the second part.
The second integral is: \[ \int \frac{3}{x^2 + 1} \, dx \] This is also a standard integral, and we know that: \[ \int \frac{1}{x^2 + 1} \, dx = \tan^{-1}(x) \] Thus: \[ \int \frac{3}{x^2 + 1} \, dx = 3 \tan^{-1}(x) \] Step 4: Putting it all together.
The full solution is: \[ \int \frac{2x - 3}{x^2 + 1} \, dx = \ln(x^2 + 1) - 3 \tan^{-1}(x) + C \] Step 5: Conclusion.
Thus, the integral is: \[ \ln(x^2 + 1) - 3 \tan^{-1}(x) + C \]