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a
23sin−12x−141+C
b
32sin−12x−341+C
c
141sin−12x−341+C
d
sin−12x−341+C
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Correct option is D
LetI=∫18+3x−x2dx=∫18−x2−3x+322−322dx=∫18−x−322−94dx=∫18+94−x−322dx=∫14122−x−322dxLet x−32=t⇒dx=dt∴I=∫14122−t2dt=sin−1t412+C∵∫dxa2−x2=sin−1xa=sin−1x−32412+C=sin−12x−341+C∵t=x−32Similar Questions
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