Single Variable Calculus, Chapter 7, 7.4-1, Section 7.4-1, Problem 86

If $f''(x) = x^{-2}$, $x > 0$, $f(1) = 0$ and $f(2) = 0$, find $f$
If $f''(x) = x^{-2}$, then by applying integration...

$\begin{equation} \begin{aligned} f'(x) &= \int x^{-2} dx\\ \\ &= \frac{x^{-1}}{-1} + c_1\\ \\ &= -\frac{1}{x} + c_1 \end{aligned} \end{equation}$

Again, by applying integration...

$\begin{equation} \begin{aligned} f(x) &= \int \left( -\frac{1}{x} + c_1 \right) dx\\ \\ f(x) &= - \ln x + c_1 x + c_2 \end{aligned} \end{equation}$


If $f(1) = 0$, then

$\begin{equation} \begin{aligned} 0 &= - \ln (1) + c_1(1)+c_2\\ \\ 0 &= c_1 + c_2\\ \\ c_1 &= -c_2 \qquad \Longleftarrow \text{(Equation 1)} \end{aligned} \end{equation}$

Also, if $f(2) = 0$, then

$\begin{equation} \begin{aligned} 0 &= -\ln(2) + c_1(2)+c_2\\ \\ \ln(2) &= 2c_1 + c_2 \qquad \Longleftarrow \text{(Equation 2)} \end{aligned} \end{equation}$

By using Equations 1 and 2 simultaneously...

$\begin{equation} \begin{aligned} \ln (2) &= 2c_1 - c_1\\ \\ c_1 &= \ln 2 \end{aligned} \end{equation}$

Thus, $c_2 = - \ln 2$
Therefore,
$f(x) = -\ln x + x \ln (2) - \ln (2)$