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Question:

The value of the integral ∫1/20 (1+√3)/[(x+1)²(1−x)⁶]¹⁴ dx is _______.

Solution:

∫1/20(1+√3)dx[(1+x)²(1−x)⁶]¹/⁴ ⇒ ∫1/20(1+√3)dx(1+x)¹/²[(1−x)⁶]¹/⁴ ⇒ ∫1/20(1+√3)dx(1+x)²[(1−x)⁶(1+x)⁶]¹/⁴
Put (1−x)/(1+x) = t ⇒ −2dx/(1+x)² = dt
I = ∫1/3¹ (1+√3)dt√t⁶/⁴ = −(1+√3)/2 × √√t¹/3¹ = (1+√3)(√3−1) = 2.