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Question : 76 of 160
Marks:
+1,
-0
Solution:
‌‌ Let ‌(‌)‌=y‌⇒ln‌y=‌(ln(n+r)−ln(n))‌⇒ln‌y=‌(ln(1+‌))‌⇒ln‌y=ln(1+x)‌dx‌⇒‌‌y=e0‌ln(1+x)‌dx‌‌ Let ‌P=ln(1+x)‌dx‌⇒‌‌P=[x⋅ln(1+x)]01−(‌)‌dx‌⇒‌‌P=ln‌2−(‌)‌dx‌‌ Again, letl ‌+x=t‌‌‌dx=dt‌‌‌=(‌)‌dt=(1−‌)‌dt‌‌‌=[t−ln‌t]12‌‌‌=2−ln‌2−1=1−ln‌2‌∴‌‌P=ln‌2−(1−ln‌2)‌⇒‌‌P=ln‌2−1+ln‌2‌⇒‌‌P=2‌ln‌2−1‌⇒‌‌P=ln‌4−ln‌e‌⇒‌‌P=ln‌⇒‌‌y‌=e‌ln‌y‌=‌
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