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Find the integral of 1/(x log x)

Evaluate the following integral:

    \[ \int \frac{dx}{x \log x}. \]

First, we add and subtract \log x in the numerator,

    \begin{align*}   \int \frac{1}{x \log x} \, dx &= \int \frac{1+\log x - \log x}{x \log x} \, dx \\  &= \int \frac{1+\log x}{x \log x} \, dx - \int \frac{\log x}{x \log x} \, dx \\  &= \int \frac{1+\log x}{x \log x} \, dx - \int \frac{1}{x} \, dx \\  &= \int \frac{1+\log x}{x \log x} \, dx - \log |x| + C. \end{align*}

For the integral on the left, let u = x \log x then du = 1 + \log x and so we have

    \begin{align*}    \int \frac{1+\log x}{x \log x} \, dx &= \int \frac{du}{u} \\  &= \log |u| + C \\  &= \log |x \log x| + C\\  &= \log |x| + \log | \log x| + C. \end{align*}

Therefore, we have

    \begin{align*}  \int \frac{dx}{x \log x} &= \int \frac{1+\log x}{x \log x} \, dx - \log |x| + C \\  &= \log |x| + \log |\log x| - \log|x| + C \\  &= \log|\log x| + C. \end{align*}


  1. Preet Malviya says:

    If we do the above integration by integration by parts we 1/x as second function and logx as first function we end up with the expression I = 1 + I , so can you tell me what’s wrong with this method ?

    • Freddy says:

      bro jus take logx=t and differentiate it therefore it becomes 1/x = dt/dx which can be written as dx/x = dt now substitute in the question we get int 1/t dt now we know int of 1/t is log(t) + c
      substitute t we get log(logx)+c

      the above method is lengthy and most prolly has the tendency to make teachers tear your answer sheets

  2. Anonymous says:

    what if you just let u = log x du = 1/x dx so then you have the integral of 1/u du, integrate and you get log u + c. subsititute to obtain: log | log x| + c

  3. chuck says:

    need help to find the integral ln(x)/(x+a). Really appreciate it if you can show the integral can be expressed in elementary functions.

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