Question

What is $\ln x$integration?

Answer 1

Integration of $\mathbf{ln}\mathbf{}\mathbf{x}\mathbf{}$:

Let,

$$\begin{gathered} t=\ln x \\ \Rightarrow e^t=e^{\ln x} \\ \Rightarrow e^t=x\left[\because e^{lnx}=x\right] \\ \Rightarrow e^{t}dt=dx \end{gathered}$$

Now,

$$\begin{gathered} \int \ln \left(x\right)dx=\int t{e}^{t}dt \\ \Rightarrow =t\left[e^t\right]-\int \frac{dt}{dt}{e}^{t}dt\left[\int v\times u\mathrm{dx}=u\int v\mathrm{dx}-\int \left(\frac{d}{\mathrm{dx}}\left(u\right)\int v\mathrm{dx}\right)\mathrm{dx}\right] \\ \Rightarrow =t\left[e^t\right]-\int e^{t}dt \\ \Rightarrow =t\left[e^t\right]-e^t+c \\ \Rightarrow =t\times e^t-e^t+c \\ \Rightarrow =e^t\left(t-1\right)+c \\ \Rightarrow =x\left(\ln x-1\right)+c\left[\mathrm{As}{e}^t=x,t=\ln x\right] \end{gathered}$$

Thus, the integration of $\mathbf{ln}\mathbf{}\mathbf{x}\mathbf{}$ is$\mathbf{\int }\mathbf{ln}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathbf{x}\left(\ln x-1\right)\mathbf{+}\mathbf{c}\mathbf{}$.