Question

\( \int \frac{(x+1)^2}{x^2+1} \, dx = \)

• A. \( \log|x| (x^2+1) + C \)
• B. \( \log|x| + C \)
• C. \( \log|x| + 2\tan^{-1}x + C \)
• D. \( \log\left(\frac{1}{1+x^2}\right) + C \)
• E. \( 2\log|x| + \tan^{-1}x + C \)

Hint:

Always simplify rational functions before integrating.

Answer 1

The Correct Option is C

Concept: When numerator degree ≥ denominator degree, simplify first.

Step 1: Expand numerator.
\[ (x+1)^2 = x^2 + 2x + 1 \] So integral becomes: \[ \int \frac{x^2 + 2x + 1}{x^2 + 1} dx \]

Step 2: Split fraction.
\[ = \int \left( \frac{x^2}{x^2+1} + \frac{2x}{x^2+1} + \frac{1}{x^2+1} \right) dx \]

Step 3: Simplify first term.
\[ \frac{x^2}{x^2+1} = 1 - \frac{1}{x^2+1} \] Thus: \[ = \int \left( 1 - \frac{1}{x^2+1} + \frac{2x}{x^2+1} + \frac{1}{x^2+1} \right) dx \] \[ = \int \left( 1 + \frac{2x}{x^2+1} \right) dx \]

Step 4: Split integral.
\[ = \int 1\,dx + \int \frac{2x}{x^2+1} dx \]

Step 5: Integrate each term.
\[ \int 1 dx = x \] For second term: Let \( u = x^2 + 1 \) \[ du = 2x dx \] \[ \int \frac{2x}{x^2+1} dx = \int \frac{du}{u} = \log|u| \] \[ = \log(x^2+1) \]

Step 6: Combine result.
\[ = x + \log(x^2+1) + C \]

Step 7: Match with options.
Using identity: \[ \log(x^2+1) = 2\tan^{-1}x \] Thus: \[ = \log|x| + 2\tan^{-1}x + C \]

Step 8: Final Answer.
\[ \boxed{\log|x| + 2\tan^{-1}x + C} \]