How do you find the integration of log x?

Apr 11, 2018

$\int \setminus \log \left(x\right) \setminus \mathrm{dx} = \frac{1}{\ln} \left(10\right) \left(x \ln \left(x\right) - x\right) + C = \frac{x}{\ln} \left(10\right) \left(\ln \left(x\right) - 1\right) + C$

Explanation:

$\int \setminus \log \left(x\right) \setminus \mathrm{dx} = \int \setminus \ln \frac{x}{\ln} \left(10\right) \setminus \mathrm{dx}$
$= \frac{1}{\ln} \left(10\right) \int \setminus \ln \left(x\right) \setminus \mathrm{dx}$
Using the integration by parts :
$\int \setminus f \left(x\right) g ' \left(x\right) \setminus \mathrm{dx} = \left[f \left(x\right) g \left(x\right)\right] - \int \setminus f ' \left(x\right) g \left(x\right) \setminus \mathrm{dx}$
There : $f \left(x\right) = \ln \left(x\right) , f ' \left(x\right) = \frac{1}{x} , g \left(x\right) = x , g ' \left(x\right) = 1$
So: $\int \setminus \log \left(x\right) \setminus \mathrm{dx} = \frac{1}{\ln} \left(10\right) \left(x \ln \left(x\right) - \int \setminus \mathrm{dx}\right)$
So:$\int \setminus \log \left(x\right) \setminus \mathrm{dx} = \frac{1}{\ln} \left(10\right) \left(x \ln \left(x\right) - x\right) + C = \frac{x}{\ln} \left(10\right) \left(\ln \left(x\right) - 1\right) + C$

In general, $\int \setminus {\log}_{\text{n}} \left(x\right) \setminus \mathrm{dx} = \frac{x}{\ln} \left(n\right) \left(\ln \left(x\right) - 1\right) + C$

n in RR""_+^* \ $\left\{1\right\}$, $C \in \mathbb{R}$

Apr 12, 2018

$\int {\log}_{10} \left(x\right) \mathrm{dx} = \frac{x \left(\ln \left(x\right) - 1\right)}{\ln \left(10\right)} + C$

Explanation:

Remember that:

${\log}_{a} \left(b\right) = {\log}_{c} \frac{b}{{\log}_{c} \left(a\right)}$

$\implies {\log}_{10} \left(x\right) = \ln \frac{x}{\ln} \left(10\right)$

We now have:

$\int \ln \left(x\right) \cdot \frac{1}{\ln \left(10\right)} \mathrm{dx}$

$\frac{1}{\ln \left(10\right)} \int \ln \left(x\right) \mathrm{dx}$

$\int u \mathrm{dv} = u v - \int v \mathrm{du}$

We let:

$u = \ln \left(x\right)$

$\mathrm{dv} = 1$

$\implies \mathrm{du} = \frac{d}{\mathrm{dx}} \left(\ln \left(x\right)\right)$

$\implies \mathrm{du} = \frac{1}{x}$

$\implies v = \int 1 \mathrm{dx}$

$\implies v = x$

$\implies \frac{1}{\ln \left(10\right)} \cdot \left[x \ln \left(x\right) - \int x \cdot \frac{1}{x} \mathrm{dx}\right]$

$\implies \frac{1}{\ln \left(10\right)} \cdot \left[x \ln \left(x\right) - \int 1 \mathrm{dx}\right]$

$\implies \frac{1}{\ln \left(10\right)} \cdot \left[x \ln \left(x\right) - x\right]$

$\implies \frac{x \ln \left(x\right) - x}{\ln \left(10\right)}$

$\implies \frac{x \left(\ln \left(x\right) - 1\right)}{\ln \left(10\right)}$ Do you $C$ why this is incomplete?

$\implies \int \log \left(x\right) \mathrm{dx} = \frac{x \left(\ln \left(x\right) - 1\right)}{\ln \left(10\right)} + C$