Question

Evaluate the integral \( \int \frac{x}{x^2 + 1} dx \):

• A. \( \frac{1}{2} \ln(x^2 + 1) + C \)
• B. \( \ln(x^2 + 1) + C \)
• C. \( \frac{1}{2} \tan^{-1}(x) + C \)
• D. \( \tan^{-1}(x) + C \)

Hint:

Tip: When the numerator is the derivative of the denominator (or almost), try substitution and look for the natural log form.

Answer 1

The Correct Option is A

To evaluate the integral \( \int \frac{x}{x^2 + 1} \, dx \), we employ the substitution method. Let \( u = x^2 + 1 \). Differentiating, we find \( du = 2x \, dx \), which rearranges to \( x \, dx = \frac{1}{2} du \).

Substituting these into the integral yields:

\[ \int \frac{x}{x^2 + 1} \, dx = \int \frac{1}{2} \cdot \frac{1}{u} \, du \]

This simplifies to:

\[ \frac{1}{2} \int \frac{1}{u} \, du \]

Integrating \( \frac{1}{u} \, du \) gives:

\[ \frac{1}{2} \ln|u| + C \]

Substituting \( u = x^2 + 1 \) back, we obtain:

\[ \frac{1}{2} \ln(x^2 + 1) + C \]

Thus, the integral evaluates to \( \frac{1}{2} \ln(x^2 + 1) + C \).