evaluate the following integral.\n\n int \frac { 256 } { x ^ { 3 } - 16 x ^ { 2 } } d x \n\nfind the partial…

Explanation:

Step1: Factor the denominator

Factor (x^{3}-16x^{2}=x^{2}(x - 16)).

Step2: Set up partial fraction decomposition

Let (\frac{256}{x^{2}(x - 16)}=\frac{A}{x}+\frac{B}{x^{2}}+\frac{C}{x - 16}). Multiply through by (x^{2}(x - 16)) to get (256=Ax(x - 16)+B(x - 16)+Cx^{2}).

Step3: Solve for (B)

Let (x = 0), then (256=B(-16)), so (B=-16).

Step4: Solve for (C)

Let (x = 16), then (256=C(16)^{2}), so (C = 1).

Step5: Solve for (A)

Expand (Ax(x - 16)+B(x - 16)+Cx^{2}=Ax^{2}-16Ax + Bx-16B + Cx^{2}=(A + C)x^{2}+(-16A + B)x-16B). Since (B=-16) and (C = 1), and there is no (x^{2}) term in the numerator (coefficient of (x^{2}) is (0)), (A+C=0), so (A=-1).

Answer:

(\int\frac{256}{x^{3}-16x^{2}}dx=\int\left(-\frac{1}{x}-\frac{16}{x^{2}}+\frac{1}{x - 16}\right)dx)