Question

How do you integrate the function $\ln 4x$ using integration by parts?

Answer 1

Hint: To simplify the integration, first assume $4x=t$ and convert the integration in the form of variable $t$. Then apply integration by parts assuming $\ln t$ as first function and 1 as second function. The formula to use integration by parts is $\int uvdx=u\int vdx-\int {\displaystyle \frac{du}{dx}}\left(\int vdx\right)dx$, here $u$ and $v$ are the functions of $x$.

Complete step by step answer:
According to the question, we have to integrate the function $\ln 4x$ using integration by parts.
Let this integration is denote by $I$ then we have:
$\Rightarrow I=\int \ln 4xdx.....\text{(1)}$
Now if we substitute $4x=t$ then we have:
$$\begin{gathered} \Rightarrow 4dx=dt \\ \Rightarrow dx={\displaystyle \frac{dt}{4}} \end{gathered}$$
Putting these values in the integration $I$, we have:
$$\begin{gathered} \Rightarrow I=\int \ln t{\displaystyle \frac{dt}{4}} \\ \Rightarrow I={\displaystyle \frac{1}{4}}\int \ln tdt \end{gathered}$$
The function $\ln t$ can be written as $\ln t\times 1$. By doing this we will get:
$\Rightarrow I={\displaystyle \frac{1}{4}}\int \ln t\left(1\right)dt$
Now we can use integration by parts assuming $\ln t$ as first function and 1 as second function. The formula of integration by parts is shown below:
$\Rightarrow \int uvdx=u\int vdx-\int {\displaystyle \frac{du}{dx}}\left(\int vdx\right)dx$, here $u$ and $v$ are the functions of $x$.
Using this formula for the above integration, we have:
$\Rightarrow I={\displaystyle \frac{1}{4}}\left[\ln t\int dt-\int {\displaystyle \frac{d}{dt}}\ln t\left(\int dt\right)dt\right]$
We know the formulas ${\displaystyle \frac{d}{dx}}\ln x={\displaystyle \frac{1}{x}}$ and $\int dx=x$. Using these formulas, we’ll get:
$\Rightarrow I={\displaystyle \frac{1}{4}}\left[\ln t\left(t\right)-\int {\displaystyle \frac{1}{t}}\left(t\right)dt+C\right]$, here $C$ is the integration constant.
Simplifying it further, we’ll get:
$\Rightarrow I={\displaystyle \frac{1}{4}}\left[t\ln t-\int dt\right]+{\displaystyle \frac{C}{4}}$
${\displaystyle \frac{C}{4}}$ is another constant. We will put it as $K$.
$$\begin{gathered} \Rightarrow I={\displaystyle \frac{1}{4}}\left[t\ln t-t\right]+K \\ \Rightarrow I={\displaystyle \frac{t}{4}}\left[\ln t-1\right]+K \end{gathered}$$
Putting back the value $t=4x$, we have:
$$\begin{gathered} \Rightarrow I={\displaystyle \frac{4x}{4}}\left[\ln 4x-1\right]+K \\ \Rightarrow I=x\left[\ln 4x-1\right]+K \end{gathered}$$
Putting back the value of $I$ from equation (1), we have:
 $\Rightarrow \int \ln 4xdx=x\left[\ln 4x-1\right]+K$
Thus this is the required integration.

Note: Whenever we have to integrate the multiple of two functions which cannot be integrated together while one of them can be integrated separately then we can apply integration by parts. The rule of integration by parts is given below:
$\Rightarrow \int uvdx=u\int vdx-\int {\displaystyle \frac{du}{dx}}\left(\int vdx\right)dx$, here $u$ and $v$ are the functions of $x$.
Here, we have assumed functions $u$ and $v$ cannot be integrated together. So we used integration by parts. However the function $v$ can be easily integrated which is why this method is helpful. For example in above case of integrating $\int \ln t\left(t\right)dt$, we took $\ln t$ as the first function and 1 as the second function.