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Taking tan⁻¹x as u and dx as dv, the integral is
Ax tan⁻¹x − (1/2) ln(1 + x²) + C
Bx tan⁻¹x − 2 ln(1 + x²) + C
Cx tan⁻¹x + (1/2) ln(1 + x²) + C
Dtan⁻¹x − (1/2) ln(1 + x²) + C
Answer & Solution
Correct answer: A. x tan⁻¹x − (1/2) ln(1 + x²) + C
Taking u = tan⁻¹x and dv = dx gives x tan⁻¹x − ∫x/(1 + x²)dx = x tan⁻¹x − (1/2)ln(1 + x²).
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