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If In = ∫(lnx)^ndx, then In + nIn - 1 is equal to (A) x (ln x)^n –1

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If In = ∫(lnx)ndx, then In + nIn - 1 is equal to

(A) x (ln x)n –1

(B) x (ln x)n

(C) nx (ln x)n

(D) None of these

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Answer is (B) x(lnx)n

In = ∫(lnx)ndx

In = x(lnx)n – n∫(lnx)n - 1dx

In = x(lnx)n – nIn – 1

In + nIn –1 = x(lnx)n

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