Question:
\( \int_{0}^{1} \frac{dx}{x^2+2x+2} \) is equal to
\( \int_{0}^{1} \frac{dx}{x^2+2x+2} \) is equal to
Show Hint
Use completing square for quadratic integrals.
Updated On: May 1, 2026
- \( 0 \)
- \( \frac{\pi}{4} \)
- \( -\frac{\pi}{4} \)
- \( \frac{\pi}{2} \)
- \( -\frac{\pi}{2} \)
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The Correct Option is B
Solution and Explanation
Concept: Complete square.
Step 1: Rewrite denominator.
\[ x^2+2x+2 = (x+1)^2 +1 \]
Step 2: Substitute \( t=x+1 \).
Step 3: Integral becomes.
\[ \int \frac{dt}{t^2+1} \]
Step 4: Use formula.
\[ \int \frac{1}{t^2+1} dt = \tan^{-1} t \]
Step 5: Apply limits.
\[ = \frac{\pi}{4} \]
Step 1: Rewrite denominator.
\[ x^2+2x+2 = (x+1)^2 +1 \]
Step 2: Substitute \( t=x+1 \).
Step 3: Integral becomes.
\[ \int \frac{dt}{t^2+1} \]
Step 4: Use formula.
\[ \int \frac{1}{t^2+1} dt = \tan^{-1} t \]
Step 5: Apply limits.
\[ = \frac{\pi}{4} \]
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