자연계열 3번

$\displaystyle t\ne0$이면 $\displaystyle e^{3t}-2e^{t}+e^{-t}=e^{-t}\left(e^{4t}-2e^{2t}+1\right)=e^{-t}\left(e^{2t}-1\right)^{2}>0$이다. 따라서 함수 $\displaystyle f(t)=e^{t}\ln\left(e^{3t}-2e^{t}+e^{-t}\right)$의 정의역은 $\displaystyle (-\infty,0)\cup(0,\infty)$이다.

$\displaystyle x=e^{t}$으로 치환하면$\displaystyle \int_{\ln2}^{\ln3}f(t)\,dt=\int_{\ln2}^{\ln3}e^{t}\ln\left(e^{3t}-2e^{t}+e^{-t}\right)dt=\int_{2}^{3}\ln\left(x^{3}-2x+x^{-1}\right)dx$이다.

$\displaystyle \ln xy=\ln x+\ln y$을 이용하면$\displaystyle \begin{aligned} \int_{2}^{3}\ln\left(x^{3}-2x+x^{-1}\right)dx &=\int_{2}^{3}\ln\left\{x^{-1}\left(x^{4}-2x^{2}+1\right)\right\}dx \\ &=\int_{2}^{3}\left\{\ln\left(x^{-1}\right)+\ln\left(x^{4}-2x^{2}+1\right)\right\}dx \end{aligned}$이다. $\displaystyle \ln x^{a}=a\ln x$과 부분적분을 이용하면$\displaystyle \int_{2}^{3}\ln\left(x^{-1}\right)dx=-\int_{2}^{3}\ln x\,dx=-\Biggl[x\ln x\Biggr]_{2}^{3}+\int_{2}^{3}1\,dx=-3\ln3+2\ln2+1$$\displaystyle \begin{aligned} \int_{2}^{3}\ln\left(x^{4}-2x^{2}+1\right)dx &=\int_{2}^{3}\ln\left\{(x-1)^{2}(x+1)^{2}\right\}dx \\ &=2\int_{2}^{3}\ln(x-1)\,dx+2\int_{2}^{3}\ln(x+1)\,dx \end{aligned}$을 얻는다.

부분적분을 이용하면$\displaystyle \int_{2}^{3}\ln(x-1)\,dx=\Biggl[(x-1)\ln(x-1)\Biggr]_{2}^{3}-\int_{2}^{3}1\,dx=2\ln2-1\ln1-1=2\ln2-1$$\displaystyle \int_{2}^{3}\ln(x+1)\,dx=\Biggl[(x+1)\ln(x+1)\Biggr]_{2}^{3}-\int_{2}^{3}1\,dx=4\ln4-3\ln3-1=8\ln2-3\ln3-1$이다. 정리하면$\displaystyle \begin{aligned} \int_{2}^{3}\ln\left(x^{3}-2x+x^{-1}\right)dx &=\int_{2}^{3}\left\{\ln\left(x^{-1}\right)+\ln\left(x^{4}-2x^{2}+1\right)\right\}dx \\ &=-3\ln3+2\ln2+1+2(2\ln2-1)+2(8\ln2-3\ln3-1) \\ &=22\ln2-9\ln3-3 \end{aligned}$를 얻는다.