# What is the antiderivative `int ln x/x dx`

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The integral `int ln x/x dx` has to be determined.

let `ln x = y`

`dy/dx = 1/x`

`dy = (1/x)*dx`

`int ln x/x dx`

=> `int y dy`

=> `y^2/2 + C`

=> `(ln x)^2/2 + C`

**The integral **`int ln x/x dx = (ln x)^2/2 + C`