Question

What is antiderivative of $\ln x$?

Answer 1

Hint: Now we are given with the function $\ln x$ . Now we want to find the value of $\int{\ln xdx}$ . To evaluate the integral we will use integration by parts. Hence we will evaluate the integral by the formula $\int{udv}=uv-\int{vdu}$ . Now we will simplify the obtained expression and hence find the solution of the given problem.

Complete step-by-step solution:
Now we know that the antiderivative of function is nothing but the integration of the function.
Hence we want to find the value of integral $\int{\ln x}dx$ . Now we have no standard integral to this integral. Hence we will have to solve the integral.
We will solve this integration by using integration by parts. Now integration by parts is a general method to multiply two functions in multiplication. With the help of integration by parts we have $\int{udv}=uv-\int{vdu}$ . Now in the given integral we have $u=\ln x$ and $dv=dx$ . Hence using the formula of integration by parts we get,
$\Rightarrow \int{\ln xdx}=\ln x.x-\int{x\left( \dfrac{d\ln x}{dx} \right)}$
Now we know that the differentiation of $\ln x$ is $\dfrac{1}{x}$ . Hence substituting this we get,
$\begin{align}
  & \Rightarrow \int{\ln xdx}=x\ln x-\int{x\left( \dfrac{1}{x} \right)dx} \\
 & \Rightarrow \int{\ln xdx}=\ln x.x-\int{dx} \\
 & \Rightarrow \int{\ln xdx}=\ln x.x-x+C \\
\end{align}$
Now taking x common from the terms we get,
$\Rightarrow \int{\ln xdx}=x\left( \ln x-1 \right)+C$
Hence the antiderivative of the function $\ln x$ is nothing but $x\left( \ln x-1 \right)$.

Note: Now note that here the formula of integration by parts is given as $\int{udv}=uv-\int{vdu}$ . Now we can also write the formula as \[\int{uv}=u\int{vdv}-\int{\left( du\int{vdv} \right)}\] . Hence with the help of the formula we can find the integration of any two functions in multiplication.