【数学Ⅲ】積分計算の型網羅part3(分数関数）

【例】

\(\displaystyle\frac{1}{(x-3)(x-7)}=\frac{1}{4}\cdot\left(\displaystyle\frac{1}{x-7}-\frac{1}{x-3}\right)\)

\(\displaystyle\frac{1}{x},\quad \frac{2x+3}{x^2+3x},\quad \frac{\cos x}{\sin x}\)

【計算例】

\(\displaystyle\int \frac{2x+3}{x^2+3x}dx =\log|x^2+3x+C|\)

\(\displaystyle\int \displaystyle\frac{6x+6}{3x^2+6x+7} dx=\log (3x^2 +6x+7)+C\)

\(\displaystyle\int \frac{e^x-e^{-x}}{e^x+e^{-x}} dx =\log (e^x+e^{-x})+C\)

【計算例】

\(\displaystyle\int \frac{4}{2x+3} dx =2 \log |2x+3|+C\)

\(\displaystyle\int \frac{4}{7x+9} dx =\frac{4}{7} \log |7x+9|+C\)

\(\displaystyle\frac{1}{x^2-4x-32}=\frac{1}{(x+4)(x-8)}\)

\(\displaystyle\frac{2}{x^2-1}=\frac{2}{(x-1)(x+1)}\)

\(\displaystyle\frac{1}{x^2-4x-32}=\frac{1}{(x+4)(x-8)}=\frac{1}{12}\left(\frac{1}{x-8}-\frac{1}{x+4}\right)\)

\(\displaystyle\frac{2}{x^2-1}=\frac{2}{(x-1)(x+1)}=\frac{1}{x-1}-\frac{1}{x+1}\)

\(\begin{aligned}\displaystyle\int \frac{1}{12}\left(\frac{1}{x-8}-\frac{1}{x+4}\right) dx &=\frac{1}{12}\left(\log|x-8|-\log |x+4|\right)\\&=\frac{1}{12}\log \left|\frac{x-8}{x+4}\right|+C\end{aligned}\)

\(\begin{aligned}\displaystyle\int \left(\frac{1}{x-1}-\frac{1}{x+1}\right) dx &=\log |x-1| -\log |x+1|\\&=\log |\frac{x-1}{x+1}|+C \end{aligned}\)

【例】\(\displaystyle\int \frac{3x+5}{(x+1)^2} dx\)

まずは部分分数分解です。

\(\displaystyle\frac{3x+5}{(x+1)^2}=\frac{A}{(x+1)^2}+\frac{B}{x+1}\)

として\(A, B\)を求めると、

\(A=2,\quad B=3\)

となるから

\(\displaystyle\frac{3x+5}{(x+1)^2}=\frac{2}{(x+1)^2}+\frac{3}{x+1}\)

これを\(x\)で積分していきましょう。

\(\begin{aligned}\displaystyle\int \frac{3x+5}{(x+1)^2} dx &=\int \left(\frac{2}{(x+1)^2}+\frac{3}{x+1}\right) dx\\&=-\frac{2}{x+1}+3\log |x+1| +C\end{aligned}\)

【例1】 \(\displaystyle\int_0^a \frac{1}{x^2 +a^2} dx\)

\(x=a\tan\theta\)と置換すると、被積分関数は

\(\displaystyle\frac{1}{x^2+a^2}=\displaystyle\frac{1}{a^2(1+\tan^2\theta)}=\displaystyle\frac{\cos^2\theta}{a^2}\)

となり、\(\displaystyle\frac{dx}{d\theta}=\displaystyle\frac{a}{\cos^2\theta}\)で\(x:0\to a\)のとき\(\theta:0\to\displaystyle\frac{\pi}{4}\)となるから

\(\begin{aligned}\displaystyle\int_{0}^{a}\displaystyle\frac{1}{x^2+a^2}dx&=\displaystyle\int_{0}^{\frac{\pi}{4}}\displaystyle\frac{\cos^2\theta}{a^2}\displaystyle\frac{a}{\cos^2\theta}d\theta\\&=\displaystyle\int_{0}^{\frac{\pi}{4}}\displaystyle\frac{1}{a}d\theta\\&=\displaystyle\frac{\pi}{4a}\end{aligned}\)

【例2】\(I=\displaystyle\int_{-1}^0 \frac{1}{x^2 +2x +2} dx\)

被積分関数の分母を\((ax+b)^2 +A\)の形にしましょう。

\(\begin{aligned}I=\displaystyle\int_{-1}^0 \frac{1}{x^2 +2x +2} dx&=\int \frac{1}{(x+1)^2+1} dx \end{aligned}\)

ここで、\(x+1=t\)と置換すると、\(dx=dt\)かつ、\(x:-1\to 0\)で\(t:0\to 1\)だから

\(\begin{aligned}I=\displaystyle\int_{-1}^0 \frac{1}{(x+1)^2+1} dx &=\int_0^1 \frac{1}{1+t^2} dt\end{aligned}\)

ここで先ほど同様、\(t=\tan \theta\)とおくと、\(\displaystyle\frac{dx}{d\theta}=\displaystyle\frac{1}{\cos^2\theta}\)で\(x:0\to 1\)のとき\(\theta:0\to\displaystyle\frac{\pi}{4}\)となるから

\(\begin{aligned}I&=\displaystyle\int_0^1 \frac{1}{1+t^2} dt\\&=\int_{0}^{\frac{\pi}{4}} d\theta \\&=\frac{\pi}{4}\end{aligned}\)

となります。