題目 5. (10 points) Evaluate the improper integral ∫1∞3x2+1x4+x2 dx.\int_{1}^{\infty}\dfrac{3x^2+1}{x^4+x^2}\,dx.∫1∞x4+x23x2+1dx. 解答 解法一 3x2+1x4+x2=1x2+2x2+1.\frac{3x^2+1}{x^4+x^2}=\frac1{x^2}+\frac2{x^2+1}.x4+x23x2+1=x21+x2+12. 故瑕積分收斂，且 ∫1∞(1x2+2x2+1) dx=1+π2.\int_1^\infty\left(\frac1{x^2}+\frac2{x^2+1}\right)\,\mathrm{d}x =\boxed{1+\frac{\pi}{2}}.∫1∞(x21+x2+12)dx=1+2π.
