If ∫ sin⁻¹(√(x/(1+x))) dx = A(x)tan⁻¹(√x) + B(x) + C, where C is a constant of integration, then the ordered pair (A(x), B(x)) can be:
If ∫ sin⁻¹(√(x/(1+x))) dx = A(x)tan⁻¹(√x) + B(x) + C, where C is a constant of integration, then the ordered pair (A(x), B(x)) can be:
Option 1 -
(x − 1, -√x)
Option 2 -
(x − 1, √x)
Option 3 -
(x + 1, −√x)
Option 4 -
(x + 1, √x)
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1 Answer
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Correct Option - 3
Detailed Solution:I = ∫sin? ¹ (√x/√1+x)dx
∫tan? ¹ (√x)dx
= xtan? ¹√x - ∫ (1/ (1+x) * 1/ (2√x)xdx + C= xtan? ¹√x - ∫ (t²/ (1+t²) * (t*2t dt)/ (2t) + C (x=t²)
= xtan? ¹√x - ∫ (t²/ (1+t²)dt + C = xtan? ¹√x - t + tan? ¹t + C = xtan? ¹√x - √x + tan? ¹√x + C
= (x+1)tan? ¹√x - √x + C => (Ax) = x+1 => B (x) = -√x
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