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Question : 78 of 160
Marks:
+1,
-0
Solution:
Let
I=dx Let
log‌x=t ∴ Upper limit
t=log‌=−1 ⇒dx=dt lower limit
t=log‌ =log‌2−log‌e=log‌2−1 ∴I= =|−1log‌2−1 =((−1)2∕3−(log‌2−1)2∕3) I=[1−(log‌2−1)2∕3]
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