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Q.

∫2−3x−4x2dx

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a

1168x+32−3x−4x2+4164sin−18x+341+C

b

1168x+32−3x−4x2−4164sin−18x+341+C

c

1168x+32−3x−4x2+4164cos−18x+341+C

d

1168x+32−3x−4x2−4164cos−18x+341+C

answer is A.

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Detailed Solution

2∫4182−x+382dx Using ∫a2−x2dx=x2a2−x2+a22sin−1⁡xa+C=1168x+32−3x−4x2+4164sin−18x+341+C
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∫2−3x−4x2dx