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An Intiative by Sri Chaitanya
a
4π[xSin–1x+1–x2]−x+c
b
1π[xSin–1x+1–x2]−x+c
c
2π[xSin–1x+1–x2]−x+c
d
2π[xSin–1x–1–x2]−x+c
answer is A.
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