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Question: How do you integrate \(\int{4{{e}^{x}}\ln x dx}\) using integration by parts?...

How do you integrate ∫4exln⁡xdx\int{4{{e}^{x}}\ln x dx} using integration by parts?

Explanation

Solution

To integrate the given integration using integration by parts, we are going to use the following integration by parts formula which is equal to: ∫f(x)g(x)dx=f(x)∫g(x)dx−∫f′(x)∫g(x)dx\int{f\left( x \right)g\left( x \right)dx}=f\left( x \right)\int{g\left( x \right)dx}-\int{f'\left( x \right)}\int{g\left( x \right)dx}. The catch in this integration is that what could be the first term and what could be the second term. And this priority of the function is decided by the mnemonics named “ILATE” and we will discuss each letter in the below solution.

Complete step by step solution:
The integration which we have to find is as follows:
∫4exln⁡xdx\int{4{{e}^{x}}\ln x dx}
Now, we are going to integrate by using integration by parts. The formula for integration by parts is as follows:
∫f(x)g(x)dx=f(x)∫g(x)dx−∫f′(x)∫g(x)dx\int{f\left( x \right)g\left( x \right)dx}=f\left( x \right)\int{g\left( x \right)dx}-\int{f'\left( x \right)}\int{g\left( x \right)dx} …………… (1)
In the above formula, f(x)&g(x)f\left( x \right)\And g\left( x \right) are two functions but here, we should know which function will get the priority over another. To decide the priority there is mnemonic which is equal to:
ILATE
Now, in the above mnemonic, “I” represents inverse functions, “L” represents logarithmic functions, “A” represents algebraic functions, “T” represents trigonometric functions and “E” represents exponential functions. Now, the order of the letters in this word “ILATE” will be the order of priority for the functions.
So, using this mnemonic in the given integration, in the given integration, we have two functions, exponential and logarithmic so logarithmic will get priority over exponential functions. Now, f(x)=log⁡x&g(x)=exf\left( x \right)=\log x\And g\left( x \right)={{e}^{x}} then substituting these values of f(x) and g(x) in eq. (1) we get,
∫log⁡xexdx=log⁡x∫exdx−∫(log⁡x)′∫exdx\int{\log x{{e}^{x}}dx}=\log x\int{{{e}^{x}}dx}-\int{\left( \log x \right)'}\int{{{e}^{x}}dx}
Multiplying 4 on both the sides of the above equation we get,
4∫log⁡xexdx=4(log⁡x∫exdx−∫(log⁡x)′∫exdx)4\int{\log x{{e}^{x}}dx}=4\left( \log x\int{{{e}^{x}}dx}-\int{\left( \log x \right)'}\int{{{e}^{x}}dx} \right)
We know that differentiation of log⁡x\log x with respect to x is 1x\dfrac{1}{x} and also integration of ex{{e}^{x}} with respect to x is ex{{e}^{x}}so applying these integration and differentiation in the above equation we get,
4∫log⁡xexdx=4(log⁡xex−∫1xexdx)4\int{\log x{{e}^{x}}dx}=4\left( \log x{{e}^{x}}-\int{\dfrac{1}{x}}{{e}^{x}}dx \right)
Now, we are going to put the value of integration of ∫1xexdx\int{\dfrac{1}{x}{{e}^{x}}dx} as follows:
We know that:
∫exxdx=Ei(x)\int{\dfrac{{{e}^{x}}}{x}dx}=Ei\left( x \right)
In the above equation, Ei(x)Ei\left( x \right) is equal to exponential integral function so we can substitute ∫1xexdx\int{\dfrac{1}{x}{{e}^{x}}dx} as Ei(x) in the above equation and we get,
⇒4∫log⁡xexdx=4(log⁡xex−Ei(x))\Rightarrow 4\int{\log x{{e}^{x}}dx}=4\left( \log x{{e}^{x}}-Ei\left( x \right) \right)
Hence, the value of integration of the above integration is equal to 4(log⁡xex−Ei(x))4\left( \log x{{e}^{x}}-Ei\left( x \right) \right).

Note: The point to be noted is that while using integration by parts in solving the above problem, make sure to find the priority of the functions first then proceed because sometimes, in question, the first function given may or may not be the one which is following “ILATE” mnemonics.