\\(\\int\\frac{6x^{2}-29x - 10}{x^{3}-7x^{2}+8x + 16}dx\\)

Explanation:

Step1: Factor the denominator

First, factor (x^{3}-7x^{2}+8x + 16). By trial - and - error, we find that (x = 4) is a root. Using synthetic division or long - division, we get (x^{3}-7x^{2}+8x + 16=(x - 4)(x^{2}-3x - 4)=(x - 4)(x - 4)(x+1)=(x - 4)^{2}(x + 1)).

Step2: Decompose into partial fractions

Let (\frac{6x^{2}-29x - 10}{x^{3}-7x^{2}+8x + 16}=\frac{A}{x + 1}+\frac{B}{x - 4}+\frac{C}{(x - 4)^{2}}). Then (6x^{2}-29x - 10=A(x - 4)^{2}+B(x + 1)(x - 4)+C(x + 1)). If (x=-1), then (6+29 - 10=A(-1 - 4)^{2}), (25 = 25A), so (A = 1). If (x = 4), then (6\times16-29\times4-10=C(4 + 1)), (96-116 - 10 = 5C), (-30=5C), so (C=-6). Expand the right - hand side: (A(x - 4)^{2}+B(x + 1)(x - 4)+C(x + 1)=A(x^{2}-8x + 16)+B(x^{2}-3x - 4)+C(x + 1)=(A + B)x^{2}+(-8A-3B + C)x+(16A-4B + C)). Since (A = 1) and (C=-6), and the coefficient of (x^{2}) is (6), we have (A + B=6), so (B = 5). So (\frac{6x^{2}-29x - 10}{x^{3}-7x^{2}+8x + 16}=\frac{1}{x + 1}+\frac{5}{x - 4}-\frac{6}{(x - 4)^{2}}).

Step3: Integrate term - by - term

(\int\frac{6x^{2}-29x - 10}{x^{3}-7x^{2}+8x + 16}dx=\int\frac{1}{x + 1}dx+\int\frac{5}{x - 4}dx-\int\frac{6}{(x - 4)^{2}}dx). (\int\frac{1}{x + 1}dx=\ln|x + 1|), (\int\frac{5}{x - 4}dx=5\ln|x - 4|), (\int\frac{6}{(x - 4)^{2}}dx=6\int(x - 4)^{-2}dx=-\frac{6}{x - 4}).

Answer:

(\ln|x + 1|+5\ln|x - 4|+\frac{6}{x - 4}+C)