Question
If ∫dxx√1−x3=a log∣∣∣√1−x3−1√1−x3+1∣∣∣+C then a =
Answer: Option A
:
A
Put 1−x3=t2⇒−3x2dx=2tdt∴∫dxx√1−x3=∫x2x3√1−x3dx=23∫dtt2−1=13log∣∣t−1t+1∣∣+C=13log∣∣∣√1−x3−1√1−x3+1∣∣∣+C∴a=13
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:
A
Put 1−x3=t2⇒−3x2dx=2tdt∴∫dxx√1−x3=∫x2x3√1−x3dx=23∫dtt2−1=13log∣∣t−1t+1∣∣+C=13log∣∣∣√1−x3−1√1−x3+1∣∣∣+C∴a=13
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