微積3.2.2

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問題3.2.2

 次の不定積分を求めよ。

(1)$\displaystyle \int \dfrac{1}{x\sqrt{x+1}} dx$

(2)$\displaystyle \int \dfrac{x\sqrt x}{1+\sqrt x} dx$

(3)$\displaystyle \int \dfrac{x^2}{\sqrt{x^2+1}} dx$

(4)$\displaystyle \int \dfrac{1}{(x+1)^2 \sqrt{1-x^2}} dx$

 

《ポイント》

ややハイレベルな置換積分の問題です。

 


 

《解答例》

(1)

$\displaystyle \int \dfrac{1}{x\sqrt{x+1}} dx$

$\sqrt{x+1}=t$と置くと、$dx=2tdt$であるから、

$\begin{align}& \ \ \ \ \ \displaystyle \int \dfrac{1}{x\sqrt{x+1}} dx \\
&=\displaystyle \int \dfrac{1}{(t^2-1)t} \cdot 2tdt \\
&=2\displaystyle \int \dfrac{1}{t^2-1} dt=\displaystyle \int \left( \dfrac{1}{t-1}-\dfrac{1}{t+1} \right) dt \\
&= \log ⁡|t-1|- \log ⁡|t+1|+C \\
&= \log \left| \dfrac{t-1}{t+1} \right|+C \\
&= \log ⁡\left| \dfrac{\sqrt{x+1}-1}{\sqrt{x+1}+1} \right|+C \ \ \cdots \cdots \text{(答)} \end{align}$

 

(2)

$\displaystyle \int \dfrac{x\sqrt x}{1+\sqrt x} dx$

$\sqrt x=t$と置くと、$dx=2tdt$であるから、

$\begin{align}& \ \ \ \ \ \displaystyle \int \dfrac{x\sqrt x}{1+\sqrt x} dx \\
&=\displaystyle \int \dfrac{t^3}{1+t} \cdot 2tdt \\
&=2\displaystyle \int \dfrac{t^4}{1+t} dt \\
&=2\displaystyle \int \left( t^3-t^2+t-1+\dfrac{1}{t+1} \right) dt \\
&=2\left( \frac{1}{4} t^4-\dfrac{1}{3} t^3+\dfrac{1}{2} t^2-t+ \log ⁡|t+1| \right)+C \\
&=\dfrac{1}{2} x^2-\dfrac{2}{3} x\sqrt x+x-2\sqrt x+2 \log (\sqrt x+1)+C \ \ \cdots \cdots \text{(答)} \end{align}$

 

(3)

$\displaystyle \int \dfrac{x^2}{\sqrt{x^2+1}} dx$

$\sqrt{x^2+1}=t-x$と置くと、$x=\dfrac{t^2-1}{2t}$ より、$\sqrt{x^2+1}=\dfrac{t^2+1}{2t}$、$dx=\dfrac{t^2+1}{2t^2}dt$であるから、

$\begin{align}& \ \ \ \ \ \displaystyle \int \dfrac{x^2}{\sqrt{x^2+1}} dx \\
&=\displaystyle \int \dfrac{\left(\dfrac{t^2-1}{2t}\right)^2}{\dfrac{t^2+1}{2t}} \cdot \dfrac{t^2+1}{2t^2} dt \\
&=\displaystyle \int \dfrac{t^4-2t^2+1}{4t^3} dt=\displaystyle \int \left( \frac{1}{4} t-\dfrac{1}{2}t+\dfrac{1}{4t^3}\right) dt \\
&=\dfrac{1}{8} t^2-\dfrac{1}{2} \log ⁡|t|-\dfrac{1}{8t^2}+C \\
&=\dfrac{1}{8} \left(t^2-\dfrac{1}{t^2}\right)-\dfrac{1}{2} \log ⁡|t|+C \ \ \cdots \cdots \text{(答)} \end{align}$

ここで、$\dfrac{1}{t}=\dfrac{1}{\sqrt{x^2+1}+x}=\sqrt{x^2+1}-x$より、

$\begin{align}& \ \ \ \ \ t^2-\dfrac{1}{t}^2 \\
&=(\sqrt{x^2+1}+x)^2-(\sqrt{x^2+1}-x)^2 \\
&=4x\sqrt{x^2+1} \end{align}$

であるから、

$\begin{align}& \ \ \ \ \ \dfrac{1}{8} \left(t^2-\dfrac{1}{t^2}\right)-\dfrac{1}{2} \log ⁡|t|+C \\
&=\dfrac{1}{2} x\sqrt{x^2+1}-\dfrac{1}{2} \log ⁡|\sqrt{x^2+1}+x|+C \ \ \cdots \cdots \text{(答)} \end{align}$

 

〈別解〉(部分積分を用いる解法)

$\displaystyle \int \dfrac{x^2}{\sqrt{x^2+1}} dx$ ($=I$ と置く)

$\begin{align}&=\displaystyle \int x \cdot \dfrac{x}{\sqrt{x^2+1}} dx \\
&=\displaystyle \int x \cdot (\sqrt{x^2+1})’ dx \\
&=x\sqrt{x^2+1}-\displaystyle \int \sqrt{x^2+1} dx+C’ \ \ \cdots \cdots \text{①} \end{align}$
ここで、

$\begin{align}& \ \ \ \ \ \displaystyle \int \sqrt{x^2+1} dx \\
&=\displaystyle \int \dfrac{x^2+1}{\sqrt{x^2+1}} dx \\
&=\displaystyle \int \left( \dfrac{x^2}{\sqrt{x^2+1}}+\dfrac{1}{\sqrt{x^2+1}} \right) dx \\
&=I+\displaystyle \int \dfrac{1}{\sqrt{x^2+1}} dx \ \ \cdots \cdots \text{②} \end{align}$

$(\ast) \ \ x= \tan ⁡ \theta $と置くと、$dx=\dfrac{1}{\cos^2 ⁡ \theta} d \theta$ であるから、

$\begin{align}& \ \ \ \ \ \displaystyle \int \dfrac{1}{\sqrt{x^2+1}} dx \\
&=\displaystyle \int \dfrac{1}{\sqrt{\tan ^2⁡ \theta +1}} \cdot \dfrac{1}{\cos^2 ⁡ \theta} d \theta \\
&=\displaystyle \int \dfrac{1}{\cos ⁡ \theta }d \theta \\
&=\displaystyle \int \dfrac{\cos ⁡ \theta }{\cos^{2}⁡ \theta d \theta} \\
&=\displaystyle \int \dfrac{\cos ⁡ \theta}{1- \sin^{2}⁡ \theta } d \theta \\
&=\dfrac{1}{2} \displaystyle \int \left( \dfrac{\cos ⁡ \theta}{1+\sin \theta }+ \dfrac{\cos ⁡ \theta}{1- \sin \theta } \right) d \theta \\
&=\dfrac{1}{2} ( \log ⁡|1+ \sin ⁡ \theta |- \log ⁡|1- \sin ⁡ \theta | )+C'{}’ \\
&=\dfrac{1}{2} \log \left| \dfrac{1+ \sin ⁡ \theta }{1- \sin ⁡ \theta } \right|+C'{}’ \\
&=\dfrac{1}{2} \log \left| \dfrac{1+\sqrt{\dfrac{x^2}{x^2+1}}}{1-\sqrt{\dfrac{x^2}{x^2+1}}}\right|+C'{}’ \\
&\left( ∵ \sin ⁡ \theta =\sqrt{\dfrac{x^2}{x^2+1}} \right) \\
&= \log |\sqrt{x^2+1}+x|+C'{}’ \end{align}$

これと①、②より、

$I=x\sqrt{x^2+1}-(I+ \log |\sqrt{x^2+1}+x|)+C$

$∴I=\dfrac{1}{2} \left( x\sqrt{x^2+1}- \log |\sqrt{x^2+1}+x| \right)+C \ \ \cdots \cdots \text{(答)} $

 

(注) 別解において$(\ast)$の段階で、

$x= \tan ⁡ \dfrac{\theta}{2}$ と置くと、$dx=\dfrac{1}{2 \cos^{2}⁡ \dfrac{\theta}{2}} d \theta$ であるから、

$\begin{align}& \ \ \ \ \ \displaystyle \int \dfrac{1}{\sqrt{x^2+1}} dx \\ &=\displaystyle \int \dfrac{1}{\sqrt{\tan^2⁡ \dfrac{\theta}{2}+1}} \cdot \dfrac{1}{2 \cos^{2}⁡ \dfrac{\theta}{2}}d \theta \\ &=\dfrac{1}{2} \displaystyle \int \dfrac{1}{\cos⁡ \dfrac{\theta}{2}} d \theta \\ &=\displaystyle \int \dfrac{\cos⁡ \dfrac{\theta}{2}}{\cos^2 \dfrac{\theta}{2}} d \theta \\ &=\displaystyle \int \dfrac{\cos ⁡\dfrac{\theta}{2}}{1- \sin^{2}⁡ \dfrac{\theta}{2}} d \theta \end{align}$

$\begin{align}&=\dfrac{1}{2} \displaystyle \int \dfrac{1}{2} \left( \dfrac{\cos ⁡ \dfrac{\theta}{2}}{1+ \sin ⁡ \dfrac{\theta}{2}}+\dfrac{\cos ⁡ \dfrac{\theta}{2} }{1- \sin ⁡ \dfrac{\theta}{2}} \right)d \theta \\ &=\frac{1}{4} \left(2 \log \left| 1+ \sin ⁡ \dfrac{\theta}{2} \right|-2 \log \left|1- \sin ⁡ \dfrac{\theta}{2} \right| \right) \\ &=\dfrac{1}{2} \log \left| \dfrac{1+ \sin ⁡ \dfrac{\theta}{2}}{1- \sin ⁡ \dfrac{\theta}{2}}\right| \\ &= \log |\sqrt{x^2+1}+x| \end{align}$

となり、以下同様。

 

(4)

$\displaystyle \int \dfrac{1}{(x+1)^2 \sqrt{1-x^2}} dx$

$1-x^2=0$の2実数解は$\pm 1$であるから、

$t=\sqrt{\dfrac{-(x-1)}{x+1}}$ と置くと、$x=-\dfrac{t^2-1}{t^2+1}=-1+\dfrac{2}{t^2+1}$となる。よって$dx=-\dfrac{4t}{(t^2+1)^2} dt$

$\begin{align}& \ \ \ \ \ \displaystyle \int \dfrac{1}{(x+1)^2 \sqrt{1-x^2}} dx \\
&=\displaystyle \int \dfrac{1}{\left(\dfrac{2}{t^2+1}\right)^2 \sqrt{\dfrac{2t}{t^2+1}}} \cdot \dfrac{-4t}{(t^2+1)^2} \\
&=\displaystyle \int \dfrac{1}{\left(\dfrac{2}{t^2+1}\right)^2 \cdot \dfrac{2t}{t^2+1}} \cdot \dfrac{-4t}{(t^2+1)^2} dt \end{align}$

$\begin{align}&=-\displaystyle \int \dfrac{1}{2} (t^2+1)dt \\
&=-\dfrac{1}{2} \left( \dfrac{1}{3} t^2+1 \right) t+C \\
&=-\dfrac{1}{6} \left( \dfrac{1-x}{x+1}+3 \right) \sqrt{ \dfrac{1-x}{x+1}}+C \\
&=-\dfrac{1}{3} \sqrt{ \dfrac{1-x}{x+1}} \left(\dfrac{x+2}{x+1}\right)+C \ \ \cdots \cdots \text{(答)} \end{align}$

 


 

復習例題未設定

 


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