Hi. I am trying to solve dy/dx = yLnx/x, and to solve for y, I need to integrate Ln x over x. How can I do this?
Hello
Not sure if I get that right
You want to solve
$\displaystyle \int \frac{ln(x)}{x}dx$ ?
Using the product rule leads to ( u' = 1/x and v = ln(x) )
$\displaystyle \int \frac{ln(x)}{x}dx = ln(x) ln(x) - \int \frac{1}{x} ln(x) dx$
Add $\displaystyle \int \frac{1}{x} ln(x)$ dx to both sides
$\displaystyle 2 \int \frac{ln(x)}{x}dx = ln(x) ln(x)$
devide by 2
$\displaystyle \int \frac{ln(x)}{x}dx = 0.5 * ln(x) ln(x) + C $
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Yours
Rapha