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'll use the technique of substitution. Let \( u = x^2 + 1 \). Then the derivative is \( du = 2x \, dx \), which implies \( x \, dx = \frac{1}{2} du \).

Substituting into the integral, we get:

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

This simplifies to:

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

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

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

Substituting back \( u = x^2 + 1 \), the expression becomes:

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

Therefore, the correct answer is \( \frac{1}{2} \ln(x^2 + 1) + C \).

Answer 2

Step 1: Use substitution method. 
Let \( u = x^2 + 1 \Rightarrow \frac{du}{dx} = 2x \Rightarrow du = 2x \, dx \Rightarrow x \, dx = \frac{1}{2} du \)

Step 2: Substitute in the integral. 
\[ \int \frac{x}{x^2 + 1} dx = \int \frac{1}{u} \cdot \frac{1}{2} du = \frac{1}{2} \int \frac{1}{u} du \]

Step 3: Integrate. 
\[ \frac{1}{2} \ln |u| + C = \frac{1}{2} \ln(x^2 + 1) + C \]

Step 4: Final Answer.  \( \boxed{ \frac{1}{2} \ln(x^2 + 1) + C } \)