How do you find the integral $\frac{ln x}{x}$ $ln x / x$ ?

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Answer 1 · 2015-09-12T19:08:24

$\int \frac{ln x}{x} = \frac{{(ln x)}^{2}}{2} + C$ $int ln x / x = (ln x)^2/2 + C$

$\int \frac{ln x}{x} = ?$ $int ln x / x = ?$

This is a classic application of u -substitution.

Let $u = ln x$ $u = ln x$ . Then $d u = \frac{1}{x} d x$ $du = 1/x dx$ .

Now we have

$\int \frac{ln x}{x} d x = \int u d u$ $int ln x/x dx = int u du$

We can evaluate this easily by using the power rule:

$\int u d u = \frac{u^{2}}{2} + C$ $int u du = u^2 /2 + C$

Now, substituting back $u$ $u$ , we find

$\frac{u^{2}}{2} + C = \frac{{(ln x)}^{2}}{2} + C$ $u^2 /2 + C = (ln x)^2/2 + C$

And there is our solution.

$\int \frac{ln x}{x} = \frac{{(ln x)}^{2}}{2} + C$ $int ln x / x = (ln x)^2/2 + C$

How do you find the integral lnxxln x / x?