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Question: How do you evaluate the integral of \( \int {\dfrac{{\ln \left( {2x} \right)}}{{{x^2}}}} \,dx \) ?...

How do you evaluate the integral of ∫ln⁡(2x)x2 dx\int {\dfrac{{\ln \left( {2x} \right)}}{{{x^2}}}} \,dx ?

Explanation

Solution

Hint : In the given question, we are required to find the value of the integral given to us in the question. We will be using the integration by parts method to evaluate the given integral ∫ln⁡(2x)x2 dx\int {\dfrac{{\ln \left( {2x} \right)}}{{{x^2}}}} \,dx . So, we consider the given integral as a new variable and start solving the integral given in the problem.

Complete step by step solution:
Consider I=∫ln⁡(2x)x2 dxI = \int {\dfrac{{\ln \left( {2x} \right)}}{{{x^2}}}} \,dx
In integration by parts method, we integrate a function which is a product of two functions using a formula:
∫f(x)g(x)dx=f(x)∫g(x)dx−∫[ddxf(x)∫g(x)dx]dx\int {f\left( x \right)g\left( x \right)dx = f\left( x \right)\int {g\left( x \right)dx - \int {\left[ {\dfrac{d}{{dx}}f\left( x \right)\int {g\left( x \right)dx} } \right]dx} } }
Hence, Using integration by parts method and considering ln⁡(x)\ln \left( x \right)as first function and (1x2)\left( {\dfrac{1}{{{x^2}}}} \right) as second function, we get
I=[ln⁡(x)∫1x2dx]−∫[ddx(ln⁡x)∫1x2dx]dxI = \left[ {\ln \left( x \right)\int {\dfrac{1}{{{x^2}}}dx} } \right] - \int {\left[ {\dfrac{d}{{dx}}(\ln x)\int {\dfrac{1}{{{x^2}}}dx} } \right]} dx
Now, we know that the derivative of ln⁡(2x)\ln \left( {2x} \right) with respect to x is (1x)\left( {\dfrac{1}{x}} \right) . Also, we know that integral of (1x2)\left( {\dfrac{1}{{{x^2}}}} \right) with respect to x is (−1x)\left( { - \dfrac{1}{x}} \right) .
⇒I=[ln⁡(2x)(−1x)]−∫[(1x)(−1x)]dx\Rightarrow I = \left[ {\ln \left( {2x} \right)\left( { - \dfrac{1}{x}} \right)} \right] - \int {\left[ {\left( {\dfrac{1}{x}} \right)\left( { - \dfrac{1}{x}} \right)} \right]} dx
Opening brackets and simplifying the expression, we get,
⇒I=−ln⁡(2x)x−∫(−1x2) dx\Rightarrow I = \dfrac{{ - \ln \left( {2x} \right)}}{x} - \int {\left( {\dfrac{{ - 1}}{{{x^2}}}} \right)} \,dx
⇒I=−ln⁡(2x)x+∫(1x2) dx\Rightarrow I = \dfrac{{ - \ln \left( {2x} \right)}}{x} + \int {\left( {\dfrac{1}{{{x^2}}}} \right)} \,dx
We know that the integral of 1x2\dfrac{1}{{{x^2}}} with respect to x is (−1x)\left( {\dfrac{{ - 1}}{x}} \right). So, we get,
⇒I=−ln⁡(2x)x−1x\Rightarrow I = \dfrac{{ - \ln \left( {2x} \right)}}{x} - \dfrac{1}{x}
So, the value of integral I=∫ln⁡(2x)x2 dxI = \int {\dfrac{{\ln \left( {2x} \right)}}{{{x^2}}}} \,dx is
(−ln⁡(2x)x−1x)+c\left( {\dfrac{{ - \ln \left( {2x} \right)}}{x} - \dfrac{1}{x}} \right) + c, where c is any arbitrary constant.
So, the correct answer is “(−ln⁡(2x)x−1x)+c\left( {\dfrac{{ - \ln \left( {2x} \right)}}{x} - \dfrac{1}{x}} \right) + c”.

Note : Integration by parts method can be used to solve integrals of various complex functions involving product of functions within itself easily. In the given question, the use of integration by parts method once has made the process much easier and organized. However, this might not be the case every time. We may have to apply the integration by parts method multiple times in order to reach the final answer.