The value of the integral ∫dxx√1+xn is:
(where c is integration constant)
A
1nln√1+xn1−xn+c
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B
1nln∣∣∣√1+xn+1√1−xn−1∣∣∣+c
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C
1nln∣∣∣1+xn1−xn∣∣∣+c
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D
1nln∣∣∣√1+xn−1√1+xn+1∣∣∣+c
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Solution
The correct option is D1nln∣∣∣√1+xn−1√1+xn+1∣∣∣+c Let I=∫dxx√1+xn
Put 1+xn=t2 ⇒nxn−1dx=2tdt⇒nxndxx=2tdt⇒I=∫2tdtn(t2−1)t⇒I=1n∫2dtt2−1⇒I=1n(∫dtt−1−∫dtt+1)⇒I=1nln∣∣∣t−1t+1∣∣∣+c⇒I=1nln∣∣∣√1+xn−1√1+xn+1∣∣∣+c