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Question : 127 of 150
Marks:
+1,
-0
Solution:
Let
I=∫‌Put
‌‌xn=t⇒‌‌nxn−1‌dx=dt ∴‌‌I=∫‌=∫‌=‌‌∫(‌−‌)‌dt=‌[loget−loge(t+1)]+C⇒‌‌I=‌loge(‌)+C⇒‌‌I=‌loge(‌)+C
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