Question

If $f'(x) = x + \dfrac{1}{x}$ , then the value of $f(x)$ is
(1) ${x^2} + \log x + c$
(2) $\dfrac{{{x^2}}}{2} + \log x + c$
(3) $\dfrac{x}{2} - \log x + c$
(4) None of these

Answer 1

Here a real valued function on the real line is given. A real function is a process or a function that associates each element of the set ℝ and the domain of the function is also the set ℝ or a subset of ℝ. We can easily define integration and differentiation on a real valued function. First we separate the terms and we use basic formulas of integration and find the required value.

Complete step by step answer:
The given equation is $f'(x) = x + \dfrac{1}{x}$. --(i)
Here we are given the first order derivative of the function $f(x)$ . So to obtain the real value of $f(x)$ we just integrate ${f^/}(x)$ and we get the desired solution. Let us integrate both sides of the equation (i), we get
$\int {{f^/}(x)dx = \int {xdx} + \int {\dfrac{1}{x}dx} }$
We use the basic formulas of integration i.e., $\int {xdx} = \dfrac{{{x^2}}}{2}$ and $\int {\dfrac{1}{x}dx} = \log x$ , we get
$\Rightarrow f(x) = \dfrac{{{x^2}}}{2} + \log x + c$ where c is a constant of integration.

So, the correct answer is “Option 2”.

Note:
Note that each of the terms in (i) is a function of x, so we have to multiply dx with them at the time of integration. Again we know $\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}}}$ which leads to $\int {xdx = \dfrac{{{x^2}}}{2}}$ . Another important fact is that we have to assign a constant c at the end of the integration as it is an indefinite integration. We try to remember the formula $\int {\dfrac{1}{x}dx = \log x}$ , where $\log x$ is called the logarithm function. When we use indefinite integral, we also write the constant after getting integration. If we omit the constant part then we get an error.