Question

\\(\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:

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