Write 1/(x(x + 1)²) = A/x + B/(x + 1) + D/((x + 1)²)…

Step 1. Cover-up. At $x = 0$: $A = \dfrac{1}{(0 + 1)^2} = 1$. At $x = -1$: $D = \dfrac{1}{-1} = -1$.

Step 2. The middle constant. Put $x = 1$: $\dfrac{1}{1 \times 4} = A + \dfrac{B}{2} + \dfrac{D}{4}$, so $\dfrac{1}{4} = 1 + \dfrac{B}{2} - \dfrac{1}{4}$. Then $\dfrac{B}{2} = -\dfrac{1}{2}$ and $B = -1$.

Step 3. Integrate. $F(x) = \ln x - \ln(x + 1) + \dfrac{1}{x + 1}$ for $x > 0$.

Step 4. The difference. $F(2) = \ln 2 - \ln 3 + \dfrac{1}{3}$ and $F(1) = -\ln 2 + \dfrac{1}{2}$. So $F(2) - F(1) = 2\ln 2 - \ln 3 - \dfrac{1}{6} = 0.121$.