You need to evaluate the integral using the following substitution such that:
`ln x = t => 1/x dx = dt`
`int (1/x)/((ln x)^p)dx = int (dt)/(t^p)`
Using the negative power property yields:
`int (dt)/(t^p) = int (t^(-p))dt => int (t^(-p))dt = (t^(1-p))/(1-p) + c`
Substituting back `ln x` for `t` ...
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You need to evaluate the integral using the following substitution such that:
`ln x = t => 1/x dx = dt`
`int (1/x)/((ln x)^p)dx = int (dt)/(t^p)`
Using the negative power property yields:
`int (dt)/(t^p) = int (t^(-p))dt => int (t^(-p))dt = (t^(1-p))/(1-p) + c`
Substituting back `ln x` for `t` yields:
`int (1/x)/((ln x)^p)dx = ((ln x)^(1-p))/(1-p) + c`
You need to evaluate the definite integral using the fundamental theorem of calculus such that:
`int_1^2 (1/x)/((ln x)^p)dx = ((ln x)^(1-p))/(1-p)|_1^2`
`int_1^2 (1/x)/((ln x)^p)dx = ((ln 2)^(1-p))/(1-p) - ((ln 1)^(1-p))/(1-p) `
Since`ln 1 = 0 => ((ln 1)^(1-p))/(1-p) = 0`
`int_1^2 (1/x)/((ln x)^p)dx = ((ln 2)^(1-p))/(1-p)`
Hence, since the given integral is not an improper integral, but a definite integral, with finite limits of integration, you only can evaluate it such that `int_1^2 (1/x)/((ln x)^p)dx = ((ln 2)^(1-p))/(1-p).`