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a
−1ax−ax+C
b
−2aa−xx+C
c
2aa−xx+C
d
None of these
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Correct option is B
Let I=∫dxxax−x2, put x=at⇒dx=−adtt2∴ l=∫1ataat−a2t2⋅−at2dt=∫−aa⋅at−1⋅dt=−1a∫(t−1)−1/2⋅dt=−1a⋅(t−1)(−1/2)+1−(1/2)+1+C=−2at−1+C=−2aax−1+C ∵x=at⇒t=ax=−2aa−xx+CSimilar Questions
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