Q.5) If $$f^{\prime}(x)=x+\frac{1}{x}$$, then $$f(x)$$ is
(a) $$x^2+\log |x|+C$$
(b) $$\frac{x^2}{2}+\log |x|+C$$
(c ) $$\frac{x}{2}+\log |x|+C$$
(d) $$\frac{x}{2}-\log |x|+C$$

Solution:

First Find the antiderivative of $$f^{\prime}(x)$$

Step 1: Integrate the given function $$\boldsymbol{f}^{\prime}(\boldsymbol{x})$$
To find $$f(x)$$ from its derivative $$f^{\prime}(x)=x+\frac{1}{x}$$, integrate both sides with respect to $$x$$ :
$$
\int f^{\prime}(x) d x=\int\left(x+\frac{1}{x}\right) d x
$$

Step 2: Apply the sum rule for integration
The integral of a sum is the sum of the integrals:
$$
f(x)=\int x d x+\int \frac{1}{x} d x
$$

Step 3: Use the power rule and the integral of $$\frac{1}{x}$$
Use the power rule $$\int x^n d x=\frac{x^{n+1}}{n+1}$$ for the first term and the standard integral $$\int \frac{1}{x} d x=\log |x|$$ for the second term. Remember to add the constant of integration, $$C$$ :
$$
\begin{gathered}
f(x)=\frac{x^{1+1}}{1+1}+\log |x|+C \\
f(x)=\frac{x^2}{2}+\log |x|+C
\end{gathered}
$$

Answer:

The correct option is (b)

The function $$f(x)$$ is (b) $$\frac{x^2}{2}+\log |x|+C$$.