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dxx(logx)m is equal to

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a
(log⁡x)mm+C
b
(log⁡x)m−1m−1+C
c
(log⁡x)1−m1−m+C
d
(log⁡x)1−mm+C

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detailed solution

Correct option is C

Here, given integrand is not in any standard form. So,we convert it by substitution method in a standard form and then integrate it. ∫1x(log⁡x)mdx Let  log⁡x=t⇒1x=dtdx⇒dx=xdt∴∫1x(log⁡x)mdx=∫1x(t)mxdt=∫t−mdt=t−m+1−m+1+C=(log⁡x)1−m1−m+C


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