자연계열 2번

$\displaystyle \frac{3x^4+4x^3\ln x-4x^3-8x^2\ln x+3x^2+8x\ln x+4x+4}{x^2-2x+2}=\frac{3x^4-4x^3+3x^2+4x+4}{x^2-2x+2}+4x\ln x$이고$\displaystyle \frac{3x^4-4x^3+3x^2+4x+4}{x^2-2x+2}=3x^2+2x+1+\frac{2x-2}{x^2-2x+2}+\frac{4}{x^2-2x+2}$이다. 한편,$\displaystyle \int_1^2\left(3x^2+2x+1+\frac{2x-2}{x^2-2x+2}\right)dx=\Biggl[x^3+x^2+x+\ln|x^2-2x+2|\Biggr]_1^2=11+\ln2$이고,$\displaystyle \int_1^2\frac4{x^2-2x+2}\,dx=4\int_1^2\frac1{(x-1)^2+1}\,dx=4\int_0^1\frac1{x^2+1}\,dx=4\int_0^{\frac\pi4}\frac{\sec^2\theta}{\tan^2\theta+1}\,d\theta=4\int_0^{\frac\pi4}1\,d\theta=\pi$이며,$\displaystyle \int_1^2 4x\ln x\,dx=\Biggl[2x^2\ln x\Biggr]_1^2-\int_1^2 2x\,dx=8\ln2-3$이다. 따라서$\displaystyle \int_1^2\frac{3x^4+4x^3\ln x-4x^3-8x^2\ln x+3x^2+8x\ln x+4x+4}{x^2-2x+2}\,dx=8+\pi+9\ln2$이다.