Question

Evaluate: \[ \int \frac{1}{x^2 + 4x + 5}\,dx \]

• A. \(\tan^{-1}(x+2)+C\)
• B. \(\frac{1}{2}\tan^{-1}(x+2)+C\)
• C. \(\tan^{-1}(2x+4)+C\)
• D. \(\ln(x^2+4x+5)+C\)

Hint:

Always complete the square in quadratic denominators: \[ x^2+4x+5=(x+2)^2+1 \] Then apply: \[ \int \frac{1}{x^2+a^2}\,dx = \frac{1}{a}\tan^{-1}\left(\frac{x}{a}\right)+C \]

Answer 1

The Correct Option is A

Concept: To evaluate integrals of quadratic expressions in denominator, first complete the square. Recall: \[ \int \frac{1}{x^2+a^2}\,dx = \frac{1}{a}\tan^{-1}\left(\frac{x}{a}\right)+C \]

Step 1: Completing the square. Given integral: \[ I=\int \frac{1}{x^2+4x+5}\,dx \] Rewrite denominator: \[ x^2+4x+5 = (x+2)^2+1 \] Thus: \[ I = \int \frac{1}{(x+2)^2+1}\,dx \]

Step 2: Substitution. Let: \[ u=x+2 \] Then: \[ du=dx \] Integral becomes: \[ I = \int \frac{1}{u^2+1}\,du \] Using standard formula: \[ \int \frac{1}{u^2+1}\,du = \tan^{-1}u+C \] Hence: \[ I = \tan^{-1}(x+2)+C \] Final Answer: \[ \boxed{(A)\ \tan^{-1}(x+2)+C} \]