Evaluate the integral. dxx ln(6x) We substitute u = ln(6x) because its differential is du = dx, which occurs in the integral. We have the following. dxx ln(6x) = 1 u du = u du = 2u + C = 2(ln(6x)) + C solutionspile.com

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Evaluate the integral.  dxx  ln(6x)  We substitute   u = ln(6x)   because its differential is   du =       dx,   which occurs in the integral. We have the following.  dxx  ln(6x)  =  1  u   du   =   u du  =  2u + C  =  2(ln(6x)) + C  solutionspile.com

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