Question

Find $\int{\text{ }\!\![\!\!\text{ f (logx) + }{{\text{f}}^{'}}\text{(logx) }\!\!]\!\!\text{ }dx}$.
A. x f(logx) + c
B. $\dfrac{f(\log x)}{x}$ + c
C. ex f(logx) + c
D. $\dfrac{x}{f(\log x)}$ + c

Answer 1

Hint: Integration of a given expression helps in returning to the expression which was differentiated to get an integrated expression.
To integrate quadratic expression, first the expression should be modified to get squared term and the constant term.

Complete step by step answer:
Expression whose integral has to be determined is given as:
I = $\int{\text{ }\!\![\!\!\text{ f (logx) + }{{\text{f}}^{'}}\text{(logx) }\!\!]\!\!\text{ }dx}$
Let log x = a in the above expression.
If logx = a, the derivative on both sides is taken to substitute dx in expression I by da as follows:
$\begin{align}
  & \log x=a \\
 & \dfrac{1}{x}dx=da \\
 & dx=x\text{ }da
\end{align}$
As logx = a, value of x becomes x = ea.
Substituting x for a in expression I gives,
  \[\begin{align}
  & \text{I = }\int{\text{ }\!\![\!\!\text{ f (logx) + }{{\text{f}}^{\text{ }\!\!'\!\!\text{ }}}\text{(logx) }\!\!]\!\!\text{ dx}} \\
 & \text{I = }\int{\text{ }\!\![\!\!\text{ f (a) + }{{\text{f}}^{\text{ }\!\!'\!\!\text{ }}}\text{(a) }\!\!]\!\!\text{ }\!\!\times\!\!\text{ }{{\text{e}}^{\text{a}}}\text{ da}} \\
 & \text{I}=\int{\text{ }\!\![\!\!\text{ f (a) }{{\text{e}}^{a}}\text{ }\!\!]\!\!\text{ da + }\int{\text{ }\!\![\!\!\text{ }{{\text{f}}^{\text{ }\!\!'\!\!\text{ }}}\text{(a) }{{\text{e}}^{a}}\text{ }\!\!]\!\!\text{ da}}} \\
\end{align}\]
The integration of expression is done by the method named integration by parts. Under this method, first the integration of the integral component is done keeping another component constant and then the product of integration of the first component and derivative of the second component is subtracted from it.
Integrating the first part of expression I and keeping the second part constant,
\[\begin{align}
  & \text{I =}\int{\text{ }\!\![\!\!\text{ f (a) }{{\text{e}}^{\text{a}}}\text{ }\!\!]\!\!\text{ da + }\int{\text{ }\!\![\!\!\text{ }{{\text{f}}^{\text{ }\!\!'\!\!\text{ }}}\text{(a) }{{\text{e}}^{\text{a}}}\text{ }\!\!]\!\!\text{ da}}} \\
 & \text{I = }\int{{{\text{e}}^{\text{a}}}\text{ da }\!\!\times\!\!\text{ f (a) }}-\text{ }\int{\left[ \int{{{\text{e}}^{\text{a}}}\text{da }\!\!\times\!\!\text{ d }\!\!\{\!\!\text{ f (a) }\!\!\}\!\!\text{ }} \right]}\text{ + c + }\int{\text{ }\!\![\!\!\text{ }{{\text{f}}^{\text{ }\!\!'\!\!\text{ }}}\text{(a) }{{\text{e}}^{\text{a}}}\text{ }\!\!]\!\!\text{ da}} \\
 & \text{I = }{{\text{e}}^{\text{a}}}\text{ f (a) }-\text{ }\int{\left[ {{\text{e}}^{\text{a}}}\text{ }{{\text{f}}^{\text{ }\!\!'\!\!\text{ }}}\text{(a)} \right]}\text{ da + }\int{\text{ }\!\![\!\!\text{ }{{\text{f}}^{\text{ }\!\!'\!\!\text{ }}}\text{(a) }{{\text{e}}^{\text{a}}}\text{ }\!\!]\!\!\text{ da}\text{ + c}} \\
 & \text{I = }{{\text{e}}^{\text{a}}}\text{ f (a) + c} \\
\end{align}\]
Substituting the value of a for x gives,
\[\begin{align}
  & \text{I = }{{\text{e}}^{\text{a}}}\text{ f (a) + c} \\
 & \text{I = }{{\text{e}}^{\log x}}\text{ f}\text{ (logx) + c} \\
 & \text{I = x f (logx) + c} \\
\end{align}\]

So, the correct answer is “Option A”.

Note: Another name of integration is primitive or an Antiderivative.
If f(x) is a function of x, the family of all antiderivatives of f(x) is called the integral of f(x) and is presented by$\int{\text{f(x) }dx}$.
The process of finding an integral of a given function f(x) is known as the integration of f(x).
The sign $\int{\text{f(x) }dx}$ denotes integration of f(x).