# How do you integrate (lnx)/(2x)?

May 21, 2018

Is an inmediate integral. See below

#### Explanation:

We know that if $f \left(x\right) = \ln x$ then f´(x)=1/x and with the chain rule

if $f \left(x\right) = \ln \left(g \left(x\right)\right)$ then f´(x)=(g´(x))/g(x)

Then $\int \ln \frac{x}{2 x} \mathrm{dx} = \frac{1}{4} {\ln}^{2} x + C$

May 21, 2018

$\frac{1}{4} {\left(\ln x\right)}^{2} + C$.

#### Explanation:

Let, $I = \int \ln \frac{x}{2 x} \mathrm{dx} = \frac{1}{2} \int \ln x \cdot \frac{1}{x} \mathrm{dx}$.

Now, we use IBP : $\int u \cdot v ' \mathrm{dx} = u \cdot v - \int u ' \cdot v \mathrm{dx}$, with,

$u = \ln x \mathmr{and} v ' = \frac{1}{x}$.

$\Rightarrow u ' = \frac{1}{x} \mathmr{and} v = \int \frac{1}{x} \mathrm{dx} = \ln x$.

Hence, $I = \frac{1}{2} \left[\left(\ln x\right) \left(\ln x\right) - \int \left(\frac{1}{x} \cdot \ln x\right) \mathrm{dx}\right] , i . e . ,$

$I = \frac{1}{2} {\left(\ln x\right)}^{2} - \frac{1}{2} \int \ln x \cdot \frac{1}{x} \mathrm{dx} , \mathmr{and} ,$

$I = \frac{1}{2} {\left(\ln x\right)}^{2} - I$.

$\therefore 2 I = \frac{1}{2} {\left(\ln x\right)}^{2}$.

$\Rightarrow I = \frac{1}{4} {\left(\ln x\right)}^{2} + C$.