2024年考研数学三第12题

答案: $\displaystyle\frac{1}{2} \ln 3-\displaystyle\frac{1}{8} \pi$

解析:

$f(x)=\displaystyle\frac{5}{x^{4}+3 x^{2}-4}=\displaystyle\frac{5}{\left(x^{2}+4\right)\left(x^{2}-1\right)}$
$=\displaystyle\frac{5}{(x-1)(x+1)\left(x^{2}+4\right)}$
$=\displaystyle\frac{A}{x-1}+\displaystyle\frac{B}{x+1}+\displaystyle\frac{C x+D}{x^{2}+4}$
$=\displaystyle\frac{A(x+1)\left(x^{2}+4\right)+B(x-1)\left(x^{2}+4\right)+(C x+D)\left(x^{2}-1\right)}{(x-1)(x+1)\left(x^{2}+4\right)}$
$=\displaystyle\frac{(A+B+C) x^{3}+(A-B+D) x^{2}+(4 A+4 B-C) x+(4 A-4 B-D)}{(x-1)(x+1)\left(x^{2}+4\right)}$
$\therefore\left{\begin{array}{l}A+B+C=0 \ A-B+D=0 \ 4 A+4 B-C=0 \ 4 A-4 B-D=5\end{array}, \therefore\left{\left{\begin{array}{l}A=\displaystyle\frac{1}{2} \ B=-\displaystyle\frac{1}{2} \ C=0 \ D=-1\end{array}\right.\right.\right.$
$\therefore f(x)=\displaystyle\frac{1}{2} \cdot \displaystyle\frac{1}{x-1}-\displaystyle\frac{1}{2} \cdot \displaystyle\frac{1}{x+1}-\displaystyle\frac{1}{x^{2}+4}$

$$
\begin{aligned}
& I=\int_{2}^{+\infty} \frac{5}{x^{4}+3 x^{2}-4} d x \
& =\frac{1}{2} \int_{2}^{+\infty} \frac{1}{x-1} d x-\frac{1}{2} \int_{2}^{+\infty} \frac{1}{x+1} d x-\int_{2}^{+\infty} \frac{1}{x^{2}+4} d x \
& =\left.\frac{1}{2} \ln \frac{x-1}{x+1}\right|{2} ^{+\infty}-\left.\frac{1}{2} \arctan \frac{x}{2}\right|{2} ^{+\infty} \
& =0-\frac{1}{2} \ln \frac{1}{3}-\frac{1}{2} \times \frac{\pi}{2}+\frac{1}{2} \times \frac{\pi}{4}=\frac{1}{2} \ln 3-\frac{1}{8} \pi
\end{aligned}
$$