Q.Verify that $F(x)=\dfrac{x^{3}}{3}+4x$ is an anti-derivative of $f(x)=x^{2}+4$, and hence write down $\displaystyle\int (x^{2}+4)\,dx$.

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 $f(0) = -1$:

$f(0) = \frac{0}{3} + 5(0) + c = c = -1.$

Hence $c = -1$ and

$f(x) = \frac{x^3}{3} + 5x - 1.$

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 $f(1) = \dfrac{5}{2}$:

$\log 1 + \frac{1^2}{2} + C = \frac{5}{2} \;\Rightarrow\; 0 + \frac{1}{2} + C = \frac{5}{2} \;\Rightarrow\; C = 2.$

Hence $f(x) = \log x + \dfrac{x^2}{2} + 2$, so the blank is $2$.