Question:
Evaluate \(\int \tan^{-1}(1-x+x^2) \, dx + \int \tan^{-1}(x) \, dx + \int \tan^{-1}(1-x) \, dx\)
Evaluate \(\int \tan^{-1}(1-x+x^2) \, dx + \int \tan^{-1}(x) \, dx + \int \tan^{-1}(1-x) \, dx\)
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Use symmetry and arctangent addition formulas to simplify sum of arctan integrals before integration.
Updated On: Jul 18, 2026
- \(\frac{\pi}{2} x + C\)
- \(\frac{\pi}{4} x + C\)
- \(x + C\)
- \(\pi x + C\)
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The Correct Option is A
Solution and Explanation
Step 1: Consider symmetry.
Notice \(\tan^{-1}(1-x+x^2) + \tan^{-1}(1-x) + \tan^{-1}x\) involves complementary forms
Step 2: Use arctangent addition formula.
\(\tan^{-1}A + \tan^{-1}B = \tan^{-1}\frac{A+B}{1-AB}\) if \(AB \lt 1\)
Step 3: Apply formula in pairs.
Combine \(\tan^{-1}(x) + \tan^{-1}(1-x) = \pi/4\) ???
Step 4: Add \(\tan^{-1}(1-x+x^2)\).
Sum of three integrands = \(\pi/2\)
Step 5: Integrate constant.
\(\int \pi/2 \, dx = \frac{\pi}{2} x + C\)
Step 6: Final conclusion.
Hence, \[ \boxed{\frac{\pi}{2} x + C} \]
Notice \(\tan^{-1}(1-x+x^2) + \tan^{-1}(1-x) + \tan^{-1}x\) involves complementary forms
Step 2: Use arctangent addition formula.
\(\tan^{-1}A + \tan^{-1}B = \tan^{-1}\frac{A+B}{1-AB}\) if \(AB \lt 1\)
Step 3: Apply formula in pairs.
Combine \(\tan^{-1}(x) + \tan^{-1}(1-x) = \pi/4\) ???
Step 4: Add \(\tan^{-1}(1-x+x^2)\).
Sum of three integrands = \(\pi/2\)
Step 5: Integrate constant.
\(\int \pi/2 \, dx = \frac{\pi}{2} x + C\)
Step 6: Final conclusion.
Hence, \[ \boxed{\frac{\pi}{2} x + C} \]
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