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dx2+x3x2

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a
13cos−13x−15+C
b
13tan−16x−15+C
c
13sec−16x−15+C
d
13sin−16x−15+C

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detailed solution

Correct option is D

13∫dx562−x−162 using ∫dxa2−x2=sin−1⁡xa+C


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