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B
−1√5cos7+4x3(2x+1)+c
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C
−1√5sinh−17+4x3(2x+1)+c
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D
−1√5cosh−17+4x3(2x+1)+c
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Solution
The correct option is A−1√5sin−17+4x3(2x+1)+c I=∫1(2x+1)√x2−x−2dxSubstitute2x+1=1tx=1−t2tdx=−12t2dt√x2−x−2=√1−5t2−4t2tI=−∫1√1−5t2−4tdt=−∫1
⎷(3√5)2−(5t+2√5)2dt=−1√5sin−1(5t+23)+c=−1√5sin−1(7+4x3(2x+1))+c