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Question: Find \(\int {{e^x}\left[ {\dfrac{2}{x} - \dfrac{2}{{{x^2}}}} \right]dx} \) equals: (A) \(\dfrac{{{...

Find ∫ex[2x−2x2]dx\int {{e^x}\left[ {\dfrac{2}{x} - \dfrac{2}{{{x^2}}}} \right]dx} equals:
(A) exx+c\dfrac{{{e^x}}}{x} + c
(B) ex2x2+c\dfrac{{{e^x}}}{{2{x^2}}} + c
(C) 2exx+c\dfrac{{2{e^x}}}{x} + c
(D) 2exx2+c\dfrac{{2{e^x}}}{{{x^2}}} + c

Explanation

Solution

Hint : In the given question, we are required to find the integral of the function using the integration by parts method. So, we will first choose the first and second function according to the order set by the acronym ‘ILATE’. We must remember the integrals of basic functions such as exponential functions.

Complete step-by-step answer :
In the given question, we are required to find the value of the integral ∫ex[2x−2x2]dx\int {{e^x}\left[ {\dfrac{2}{x} - \dfrac{2}{{{x^2}}}} \right]dx} .
So, we consider the given integral as a new variable.
Consider I=∫ex[2x−2x2]dxI = \int {{e^x}\left[ {\dfrac{2}{x} - \dfrac{2}{{{x^2}}}} \right]dx}
Separating both the integrals, we get,
⇒I=∫2exxdx−∫2exx2dx\Rightarrow I = \int {\dfrac{{2{e^x}}}{x}dx} - \int {\dfrac{{2{e^x}}}{{{x^2}}}dx}
Taking constants out of the integrals,
⇒I=2∫exxdx−2∫exx2dx\Rightarrow I = 2\int {\dfrac{{{e^x}}}{x}dx} - 2\int {\dfrac{{{e^x}}}{{{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 in the second integral and considering ex{e^x} as first function and 1x2\dfrac{1}{{{x^2}}} as second function, we get
\Rightarrow I = 2\int {\dfrac{{{e^x}}}{x}dx} - 2\left[ {{e^x}\int {\dfrac{1}{{{x^2}}}dx - \int {\left\\{ {\dfrac{d}{{dx}}\left( {{e^x}} \right)\int {\dfrac{1}{{{x^2}}}dx} } \right\\}dx} } } \right]
Now, using the power rule of integration ∫xndx=xn+1n+1\int {{x^n}dx} = \dfrac{{{x^{n + 1}}}}{{n + 1}}. So, we get,
\Rightarrow I = 2\int {\dfrac{{{e^x}}}{x}dx} - 2\left[ {{e^x}\left( { - \dfrac{1}{x}} \right) - \int {\left\\{ { - \dfrac{{{e^x}}}{x}} \right\\}dx} } \right]
Simplifying the expression further, we get,
⇒I=2∫exxdx−2[−exx+∫exxdx]\Rightarrow I = 2\int {\dfrac{{{e^x}}}{x}dx} - 2\left[ { - \dfrac{{{e^x}}}{x} + \int {\dfrac{{{e^x}}}{x}dx} } \right]
Opening the brackets,
⇒I=2∫exxdx−2[−exx+∫exxdx]\Rightarrow I = 2\int {\dfrac{{{e^x}}}{x}dx} - 2\left[ { - \dfrac{{{e^x}}}{x} + \int {\dfrac{{{e^x}}}{x}dx} } \right]
⇒I=2∫exxdx+2exx−2∫exxdx+c\Rightarrow I = 2\int {\dfrac{{{e^x}}}{x}dx} + 2\dfrac{{{e^x}}}{x} - 2\int {\dfrac{{{e^x}}}{x}dx} + c
Cancelling the like integrals with opposite signs, we get,
⇒I=2exx+c\Rightarrow I = 2\dfrac{{{e^x}}}{x} + c, where c is any arbitrary constant.
So, the value of integral ∫ex[2x−2x2]dx\int {{e^x}\left[ {\dfrac{2}{x} - \dfrac{2}{{{x^2}}}} \right]dx} equals 2exx+c\dfrac{{2{e^x}}}{x} + c, where c is any arbitrary constant.
So, option (C) is the correct answer.
So, the correct answer is “Option C”.

Note : Integration by parts method can be used to solve integrals of various complex functions involving the 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. Care should be taken while simplifying the expression as it can change our final answer.