The integral $$16\int_1^2 \frac{dx}{x^3(x^2+2)^2}$$ is equal to

$$I = 16 \int_1^2 \frac{dx}{x^3(x^2+2)^2}$$

$$I = 16 \int_1^2 \frac{x \, dx}{x^4(x^2+2)^2}$$

Substitute $$x^2 = t$$. Then $$2x \, dx = dt \implies x \, dx = \frac{dt}{2}$$.

$$I = 16 \int_1^4 \frac{dt/2}{t^2(t+2)^2}$$

$$I = 8 \int_1^4 \frac{dt}{t^2(t+2)^2}$$

$$\frac{1}{t^2(t+2)^2} = \frac{A}{t} + \frac{B}{t^2} + \frac{C}{t+2} + \frac{D}{(t+2)^2}$$

$$1 = A t(t+2)^2 + B(t+2)^2 + C t^2(t+2) + D t^2$$

Put $$t = 0$$: $$1 = B(2)^2 \implies B = \frac{1}{4}$$

Put $$t = -2$$: $$1 = D(-2)^2 \implies D = \frac{1}{4}$$

Comparing coefficients of $$t^3$$: $$A + C = 0 \implies C = -A$$.

Comparing coefficients of $$t^2$$: $$4A + B + 2C + D = 0$$.

Substitute $$B, D$$, and $$C = -A$$:

$$4A + \frac{1}{4} - 2A + \frac{1}{4} = 0 \implies 2A = -\frac{1}{2} \implies A = -\frac{1}{4}, C = \frac{1}{4}$$

$$I = 8 \int_1^4 \left( -\frac{1}{4t} + \frac{1}{4t^2} + \frac{1}{4(t+2)} + \frac{1}{4(t+2)^2} \right) dt$$

$$I = 2 \int_1^4 \left( -\frac{1}{t} + \frac{1}{t^2} + \frac{1}{t+2} + \frac{1}{(t+2)^2} \right) dt$$

$$I = 2 \left[ -\ln|t| - \frac{1}{t} + \ln|t+2| - \frac{1}{t+2} \right]_1^4$$

$$I = 2 \left[ \left( \ln\frac{6}{4} - \left(\frac{1}{4} + \frac{1}{6}\right) \right) - \left( \ln\frac{3}{1} - \left(1 + \frac{1}{3}\right) \right) \right]$$

$$I = 2 \left[ \left( \ln\frac{3}{2} - \frac{5}{12} \right) - \left( \ln 3 - \frac{4}{3} \right) \right]$$

$$I = \frac{11}{6} - 2\ln 2 = \frac{11}{6} - \ln(2^2) = \frac{11}{6} - \log_e 4$$