## Calculus: Early Transcendentals 8th Edition

$\displaystyle \frac{1}{3}\ln|x^{3}+1|-\frac{1}{3}\ln|x^{3}+2|+C$
$I=\displaystyle \int\frac{x^{2}}{x^{6}+3x^{3}+2}dx=\quad \left[\begin{array}{ll} u=x^{3} & \\ du=3x^{2}dx & x^{2}dx=\frac{du}{3} \end{array}\right]$ $=\displaystyle \frac{1}{3}\int\frac{du}{u^{2}+3u+2}$ factor the denominator. (2)(1)=2, $\quad$ (2+1)=3.... $=\displaystyle \frac{1}{3}\int\frac{du}{(u+1)(u+2)}$ ... use the partial fraction method $\displaystyle \frac{1}{(u+1)(u+2)}$=$\displaystyle \frac{A}{u+1}$+$\displaystyle \frac{B}{u+2}$ $1=Au+2A+Bu+B$ $1=(A+B)u+(2A+B)\Rightarrow\left\{\begin{array}{l} A+B=0\\ 2A+B=1 \end{array}\right.$ Subtract the first from the second equation: $A=1, B=-1.$ $I=\displaystyle \frac{1}{3}\int\frac{du}{u+1}-\frac{1}{3}\int\frac{du}{u+2}$ Both are type 2 integrals (from the table) $=\displaystyle \frac{1}{3}\ln|u+1|-\frac{1}{3}\ln|u+2|+C$ ... bring back x $=\displaystyle \frac{1}{3}\ln|x^{3}+1|-\frac{1}{3}\ln|x^{3}+2|+C$