横浜国立大学(理系) 2024年 問題1(1)
\[定積分 \quad \int_0^{\log \sqrt{3}} \cfrac{e^{3x}+4e^{2x}+e^x}{e^{4x}+2e^{2x}+1} dx \quad を求めよ。\]
\[I=\int_0^{\log \sqrt{3}} \cfrac{e^{3x}+4e^{2x}+e^x}{e^{4x}+2e^{2x}+1} dx=\int_0^{\log \sqrt{3}} \cfrac{e^x(e^{2x}+4e^x+1)}{(e^{2x}+1)^2} dx\]
\[ e^x=t \quad とおくと \quad e^xdx=dt \qquad \begin{array}{c|c} x & 0\ \ \rightarrow \log \sqrt{3}\\ \hline t & 1\ \ \rightarrow \sqrt{3} \ \\ \end{array} \]
\[I=\int_1^{\sqrt{3}} \cfrac{t^2+4t+1}{(t^2+1)^2}dt=\int_1^{\sqrt{3}} \big(\cfrac{t^2+1}{(t^2+1)^2} + \cfrac{4t}{(t^2+1)^2}\big)dt =\int_1^{\sqrt{3}} \cfrac{1}{t^2+1}dt + \int_1^{\sqrt{3}}\cfrac{4t}{(t^2+1)^2}dt\]
\[(1)\quad I_1=\int_1^{\sqrt{3}} \cfrac{1}{t^2+1}dt \quad において\] \[ \quad t=\tan \theta \quad とおくと \quad dt=\cfrac{d\theta}{\cos ^2 \theta} \qquad \begin{array}{c|c} t & 1\ \ \rightarrow \sqrt{3} \ \\ \hline \theta & \dfrac{\pi}{4}\ \ \rightarrow \dfrac{\pi}{3} \ \\ \end{array} \] \[\quad I_1=\int_{\scriptsize{\dfrac{\pi}{4}}}^{\scriptsize{\dfrac{\pi}{3}}} \cfrac{1}{\tan ^2 \theta+1} \times \cfrac{d\theta}{\cos ^2 \theta} =\int_{\scriptsize{\dfrac{\pi}{4}}}^{\scriptsize{\dfrac{\pi}{3}}} \cos ^2 \theta \times \cfrac{d\theta}{\cos ^2 \theta} =\int_{\scriptsize{\dfrac{\pi}{4}}}^{\scriptsize{\dfrac{\pi}{3}}} d\theta =\cfrac{\pi}{3}-\cfrac{\pi}{4}=\cfrac{\pi}{12}\]
\[(2)\quad I_2=\int_1^{\sqrt{3}} \cfrac{4t}{(t^2+1)^2}dt \quad において\] \[ \quad t^2+1=u \quad とおくと \quad 2tdt=du \qquad \begin{array}{c|c} t & 1\ \ \rightarrow \sqrt{3} \ \\ \hline u & 2\ \ \rightarrow 4 \ \\ \end{array} \] \[\quad I_2=\int_2^4 \cfrac{2du}{u^2}=-2\big[\cfrac{1}{u}\big]_2^4=-2(\cfrac{1}{4}-\cfrac{1}{2})=\cfrac{1}{2}\]
$したがって \quad I=I_1+I_2=\cfrac{\pi}{12}+ \cfrac{1}{2}$
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