The integral 16 int limits_1^2 frac{d x}{x^3( | vedclass

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(D) Let $I = 16 \int \limits_1^2 \frac{dx}{x^3(x^2+2)^2}$.Multiply numerator and denominator by $x^4$ inside the integral: $I = 16 \int \limits_1^2 \f

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$\frac{11}{6}-\log _e 4$

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(D) Let $I = 16 \int \limits_1^2 \frac{dx}{x^3(x^2+2)^2}$.Multiply numerator and denominator by $x^4$ inside the integral: $I = 16 \int \limits_1^2 \frac{dx}{x^7(1 + \frac{2}{x^2})^2}$.Let $t = 1 + \frac{2}{x^2}$,then $dt = -\frac{4}{x^3} dx$,which implies $\frac{dx}{x^3} = -\frac{dt}{4}$.Also,$x^2 = \frac{2}{t-1}$,so $x^4 = \frac{4}{(t-1)^2}$.When $x=1$,$t=3$. When $x=2$,$t=1 + \frac{2}{4} = \frac{3}{2}$.Substituting these into the integral:$I = 16 \int \limits_3^{3/2} \frac{1}{x^4(1 + \frac{2}{x^2})^2} \cdot \frac{dx}{x^3} = 16 \int \limits_3^{3/2} \frac{(t-1)^2}{4} \cdot \frac{1}{t^2} \cdot (-\frac{dt}{4}) = -\int \limits_3^{3/2} \frac{(t-1)^2}{t^2} dt$.$I = \int \limits_{3/2}^3 (1 - \frac{2}{t} + \frac{1}{t^2}) dt = [t - 2 \ln|t| - \frac{1}{t}]_{3/2}^3$.$I = (3 - 2 \ln 3 - \frac{1}{3}) - (\frac{3}{2} - 2 \ln \frac{3}{2} - \frac{2}{3}) = (\frac{8}{3} - 2 \ln 3) - (\frac{5}{6} - 2 \ln \frac{3}{2})$.$I = \frac{16-5}{6} - 2 \ln(\frac{3}{3/2}) = \frac{11}{6} - 2 \ln 2 = \frac{11}{6} - \ln 4$.

The integral $16 \int \limits_1^2 \frac{d x}{x^3(x^2+2)^2}$ is equal to

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