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a
1nlogxnxn+1+c
b
1nlogxn+1xn+c
c
logxnxn+1+c
d
none of these
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Correct option is A
I=∫dxxxn+1=∫xn−1xnxn+1dx Putting xn=t so that nxn−1dx=dt,i.e. We get xn−1dx=1ndtI=∫1ndtt(t+1)=1n∫1t−1t+1dt=1n(logt−log(t+1))+C==1nlogxnxn+1+CSimilar Questions
is equal to
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