Question

How do you use substitution to integrate $\left( {\dfrac{{\ln \left( {5x} \right)}}{x}} \right) \cdot dx$?

Answer 1

Given an expression. We have to find the integral of the expression. First, we will substitute the variable for the logarithmic expression. Then, differentiate the expression with respect to x to find the value of dx. Then, find the value of integral using the power rule of integration with respect to the variable which is substituted.
Formula used:
The power rule of integration is given by:
$\int {{x^n}dx} = \dfrac{{{x^{n + 1}}}}{{n + 1}} + C$

Complete step by step solution:
We are given the expression $\int {\left( {\dfrac{{\ln \left( {5x} \right)}}{x}} \right) \cdot dx}$. First, we will substitute the numerator equal to the some variable.
$\Rightarrow \ln \left( {5x} \right) = t$
Then, we will apply the product rule of logarithms $\ln ab = \ln a + \ln b$, we get:
$\Rightarrow \ln 5 + \ln x = t$
Now, we will differentiate both sides with respect to $x$.
$\Rightarrow \dfrac{d}{{dx}}\ln 5 + \dfrac{d}{{dx}}\ln x = \dfrac{d}{{dx}}t$
On simplifying the expression, we get:
$\Rightarrow 0 + \dfrac{1}{x} = \dfrac{{dt}}{{dx}}$
$\Rightarrow dt = \dfrac{{dx}}{x}$
Now, we will substitute $\ln \left( {5x} \right) = t$ and $dt = \dfrac{{dx}}{x}$ into the given integral.
$\Rightarrow \int {\left( {\dfrac{{\ln \left( {5x} \right)}}{x}} \right) \cdot dx} = \int {t \cdot dt}$
Now, apply the power of integral to find the solution of the expression.
$\Rightarrow \int {t \cdot dt} = \dfrac{{{t^2}}}{2} + C$
Now, again replace $t$ by $\ln \left( {5x} \right)$ into the expression.
$\Rightarrow \dfrac{{{{\left( {\ln \left( {5x} \right)} \right)}^2}}}{2} + C$
Hence, the integral of the expression is equal to $\dfrac{{{{\left( {\ln \left( {5x} \right)} \right)}^2}}}{2} + C$
Additional information: We are given the integral expression. In the integral of logarithmic expression, we must be careful while applying the logarithmic properties. The integration can be done using various methods such as Integration by parts, integration by substitution, and also using the partial function.

Note: The students must note that we have first applied the product rule of logarithms to rewrite the numerator of the given expression. The product rule is given by $\ln ab = \ln a + \ln b$. Also, students must remember that we have substituted t for the numerator and then we have to differentiate the expression. Also, students must remember that we have applied power rules to integrate the expression.