∫dxxax−x2is equal to
−1ax−ax+C
−2aa−xx+C
2aa−xx+C
None of these
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+C