Q.If $f'(x) = \dfrac{1}{x} + x$ and $f(1) = \dfrac{5}{2}$, then $f(x) = \log x + \dfrac{x^2}{2} +$ ______

Integrate $f'(x)=x^2+5$ to get $f(x)=\frac{x^3}{3}+5x+c$, then use $f(0)=-1$ to find $c=-1$.

Since $f'(x) = x^2 + 5$, we recover $f(x)$ by integrating:

$f(x) = \int (x^2 + 5)\,dx = \frac{x^3}{3} + 5x + c.$

Apply the initial condition $f(0) = -1$: …

Integrating $f'(x) = \dfrac{1}{x} + x$ gives $f(x) = \log x + \dfrac{x^2}{2} + C$; applying $f(1) = \dfrac{5}{2}$ gives $C = 2$.

Integrate the derivative:

$f(x) = \int\left(\frac{1}{x} + x\right)dx = \log x + \frac{x^2}{2} + C.$

Use the initial condition $f(1) = \dfrac{5}{2}$: …