# Thread: How to integrate LN x over x.

1. ## How to integrate LN x over x.

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?

2. Hello

Originally Posted by jaijay32
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?

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

$\displaystyle \int \frac{ln(x)}{x}dx = 0.5 * ln(x) ln(x) + C$

Yours
Rapha

3. Originally Posted by jaijay32
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?
dy/dx = y{[ln x]/x}

Rearranging,

[1/y][dy]=[ln x]/[x] dx
integrate both sides,
ln y = integrate{[lnx]/x} dx
ln y = 0.5(ln x)^2 +c

Notice when you differentiate 0.5(ln x)^2 with respect to x you'll get back [ln x]/x.

I hope I helped.