14.,x> 0, m#1x (log x)

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Solution

The integral of the function is given as,

∫ 1 x ( logx ) m dx (1)

Consider, logx=t.

Differentiate with respect to x.

logx=t 1 x dx=dt

Substitute 1 x dx=dt in equation (1) and then integrate.

∫ 1 x ( logx ) m dx = ∫ dt t m = ∫ t −m = t −m+1 1−m +C

Substitute logx=t in the above equation.

∫ 1 x ( logx ) m dx = t −m+1 1−m +C = ( logx ) 1−m 1−m +C

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