How do you evaluate the integral $\int{\dfrac{1}{1-x}dx}$ from 0 to 1?

Hint: We will use the method of substitution to solve this integral. We will substitute the denominator with a variable. Then we will obtain a standard integral. We will find the value of this standard integral and then substitute the value of the variable. Then we will put the values of the limits given to obtain the required answer.

Complete answer:
We have to find the value of the following integral,
$I=\int{\dfrac{1}{1-x}dx}$
We will use the method of substitution to solve this integral. Let us substitute $u=1-x$. So, we have $du=-dx$ and hence, $dx=-du$. So, we have the following,
\[I=-\int{\dfrac{1}{u}du}\]
We know that this is a standard integral and its value is $\int{\dfrac{1}{x}dx}=\ln x$. Therefore, we have the following,
\[I=-\ln \left( u \right)\]
Now, substituting the value of this variable, we get
\[I=-\ln \left( 1-x \right)\]
We have to find the value of the integral from 0 to 1. So, now we will first put the value of the upper limit in the above equation and then subtract the equation obtained by putting in the lower limit. So, we have,
\[\begin{align}
  & I=-\ln \left( 1-1 \right)-\left( -\ln \left( 1-0 \right) \right) \\
 & \therefore I=-\ln \left( 0 \right)+\ln \left( 1 \right) \\
\end{align}\]
The value of $\ln \left( 1 \right)=0$ but the value of $\ln \left( 0 \right)$ is not defined. Hence, the given integral can be solved as an indefinite integral but not as a definite integral.

Note: We should be aware of the difference between definite integral and indefinite integral. It is also essential to understand the definition of functions and its properties. Depending on the function to be integrated, we can choose from the other methods of integration. The other methods are integration by parts and integration by using partial fractions.