Question

evaluate the integral. (remember to use absolute values where appropriate. use c for the constant of integration.)
\\(\int\frac{17}{(x - 1)(x^{2}+16)}dx\\)

Explanation:

Step1: Use partial - fraction decomposition

We write $\frac{17}{(x - 1)(x^{2}+16)}=\frac{A}{x - 1}+\frac{Bx + C}{x^{2}+16}$. Then $17=A(x^{2}+16)+(Bx + C)(x - 1)$. Let $x = 1$, we get $17=A(1 + 16)$, so $A = 1$. Expanding gives $17=Ax^{2}+16A + Bx^{2}-Bx + Cx - C=(A + B)x^{2}+(-B + C)x+(16A - C)$. Since $A = 1$, then $A + B=0$ implies $B=-1$, and $16A - C=17$ implies $C=-1$. So $\frac{17}{(x - 1)(x^{2}+16)}=\frac{1}{x - 1}+\frac{-x - 1}{x^{2}+16}=\frac{1}{x - 1}-\frac{x}{x^{2}+16}-\frac{1}{x^{2}+16}$.

Step2: Integrate term - by - term

$\int\frac{17}{(x - 1)(x^{2}+16)}dx=\int\frac{1}{x - 1}dx-\int\frac{x}{x^{2}+16}dx-\int\frac{1}{x^{2}+16}dx$. For $\int\frac{1}{x - 1}dx=\ln|x - 1|+C_1$. For $\int\frac{x}{x^{2}+16}dx$, let $u=x^{2}+16$, $du = 2xdx$, so $\int\frac{x}{x^{2}+16}dx=\frac{1}{2}\ln|x^{2}+16|+C_2$. For $\int\frac{1}{x^{2}+16}dx=\frac{1}{4}\arctan(\frac{x}{4})+C_3$.

Step3: Combine the results

$\int\frac{17}{(x - 1)(x^{2}+16)}dx=\ln|x - 1|-\frac{1}{2}\ln|x^{2}+16|-\frac{1}{4}\arctan(\frac{x}{4})+C$

Answer:

$\ln|x - 1|-\frac{1}{2}\ln|x^{2}+16|-\frac{1}{4}\arctan(\frac{x}{4})+C$