【微分積分】4-4-4 累乗根の積分｜要点まとめ
このページでは、大学数学の微分積分で扱う「累乗根の積分」について整理します。累乗根を含む関数の積分方法を、具体的な変形手順や例題を通してわかりやすく解説します。指数の取り扱いや置換のコツなど、計算を進めるうえで重要な考え方を身につけ、積分の基礎力を確実に強化しましょう。
累乗根の積分
【累乗根の有理化置換】
\(f(u,v)\)を\(u\)と\(v\)に関する有理関数とする。
(1)\(\displaystyle \int f\left(x,\sqrt{ax^2+bx+c}\right)dx\ (a>0)\)は
\(t=\sqrt{ax^2+bx+c}+\sqrt{a}x\ \ \ \)または
\(t=\sqrt{ax^2+bx+c}-\sqrt{a}x\)
(2)\(\displaystyle \int f\left(x,\sqrt{a(x-\alpha)(x-\beta)}\right)dx\ (a\neq0)\)は
\(\displaystyle t=\sqrt{a\frac{x-\alpha}{x-\beta}}\)
(3)\(\displaystyle \int f\left(x,\sqrt[n]{\frac{ax+b}{cx+d}}\right)dx\ (n\geqq2,ad-bc\neq0)\)は
\(\displaystyle t=\sqrt[n]{\frac{ax+b}{cx+d}}\)
【例題】次の不定積分を求めなさい。
(1)\(\displaystyle \int\frac{1}{\sqrt{x}+\sqrt{x+1}}dx\)
\(\displaystyle =\int(\sqrt{x+1}-\sqrt{x})dx\)
\(\displaystyle =\frac{2}{3}(x+1)^{\frac{3}{2}}-\frac{2}{3}x^{\frac{3}{2}}+C\)
\(\displaystyle =\frac{2}{3}(x+1)^{\frac{3}{2}}-\frac{2}{3}x^{\frac{3}{2}}+C\)
(2)\(\displaystyle \int\frac{1}{\sqrt{x(3-x)}}dx\)
\(\displaystyle =\int\frac{1}{\sqrt{\frac{9}{4}-(x-\frac{3}{2})^2}}dx\)
\(\displaystyle =\sin^{-1}\frac{x-\frac{3}{2}}{\frac{3}{2}}+C\)
\(\displaystyle =\sin^{-1}\frac{2x-3}{3}+C\)
\(\displaystyle =\sin^{-1}\frac{x-\frac{3}{2}}{\frac{3}{2}}+C\)
\(\displaystyle =\sin^{-1}\frac{2x-3}{3}+C\)
(3)\(\displaystyle \int\frac{1}{\sqrt{x^2+4}}dx\)
\(=\log(x+\sqrt{x^2+4})+C\)
(4)\(\displaystyle \int\frac{1}{\sqrt{x^2-8}}dx\)
\(=\log|x+\sqrt{x^2-8}|+C\)
(5)\(\displaystyle \int\frac{1}{x\sqrt{x+4}}dx\)
\(t=\sqrt{x+4}\)とおくと、\(\displaystyle dx=2tdt\)
\(\displaystyle =\int\frac{1}{(t^2-4)t}2tdt\)
\(\displaystyle =\frac{1}{2}\int\left(\frac{1}{t-2}-\frac{1}{t+2}\right)dt\)
\(\displaystyle =\frac{1}{2}\log\left|\frac{t-2}{t+2}\right|+C\)
\(\displaystyle =\frac{1}{2}\log\left|\frac{\sqrt{x+4}-2}{\sqrt{x+4}+2}\right|+C\)
\(\displaystyle =\int\frac{1}{(t^2-4)t}2tdt\)
\(\displaystyle =\frac{1}{2}\int\left(\frac{1}{t-2}-\frac{1}{t+2}\right)dt\)
\(\displaystyle =\frac{1}{2}\log\left|\frac{t-2}{t+2}\right|+C\)
\(\displaystyle =\frac{1}{2}\log\left|\frac{\sqrt{x+4}-2}{\sqrt{x+4}+2}\right|+C\)
(6)\(\displaystyle \int\frac{x}{(x+2)\sqrt{x+1}}dx\)
\(t=\sqrt{x+1}\)とおくと、\(\displaystyle dx=2tdt\)
\(\displaystyle =\int\frac{t^2-1}{(t^2+1)t}2tdt\)
\(\displaystyle =\int\left(2-\frac{4}{1+t^2}\right)dt\)
\(\displaystyle =2t-4\tan^{-1}t+C\)
\(\displaystyle =2\sqrt{x+1}-4\tan^{-1}\sqrt{x+1}+C\)
\(\displaystyle =\int\frac{t^2-1}{(t^2+1)t}2tdt\)
\(\displaystyle =\int\left(2-\frac{4}{1+t^2}\right)dt\)
\(\displaystyle =2t-4\tan^{-1}t+C\)
\(\displaystyle =2\sqrt{x+1}-4\tan^{-1}\sqrt{x+1}+C\)
(7)\(\displaystyle \int\frac{1}{\sqrt{x}+\sqrt[3]{x}}dx\)
\(x=t^6\)とおくと、\(\displaystyle dx=6t^5dt\)
\(\displaystyle =\int\frac{6t^5}{t^3+t^2}dt\)
\(\displaystyle =\int\left(t^2-t+1-\frac{1}{t+1}\right)dt\)
\(\displaystyle =2t^3-3t^2+6t-6\log|t+1|+C\)
\(\displaystyle =2\sqrt{x}-3\sqrt[3]{x}+6\sqrt[6]{x}-6\log(\sqrt[6]{x}+1)+C\)
\(\displaystyle =\int\frac{6t^5}{t^3+t^2}dt\)
\(\displaystyle =\int\left(t^2-t+1-\frac{1}{t+1}\right)dt\)
\(\displaystyle =2t^3-3t^2+6t-6\log|t+1|+C\)
\(\displaystyle =2\sqrt{x}-3\sqrt[3]{x}+6\sqrt[6]{x}-6\log(\sqrt[6]{x}+1)+C\)
(8)\(\displaystyle \int\frac{1}{x\sqrt{x^2+x-1}}dx\)
\(t=\sqrt{x^2+x-1}\)とおくと、\(\displaystyle dx=\frac{2t^2+2t-2}{(2t+1)^2}dt\)
\(\displaystyle =\int\frac{1}{\frac{t^2+1}{2t+1}・\frac{t^2+t-1}{2t+1}}・\frac{2t^2+2t-2}{(2t+1)^2}dt\)
\(\displaystyle =\int\frac{2}{t^2+1}dt\)
\(\displaystyle =2\tan^{-1}t+C\)
\(\displaystyle =2\tan^{-1}(\sqrt{x^2+x-1}+x)+C\)
\(\displaystyle =\int\frac{1}{\frac{t^2+1}{2t+1}・\frac{t^2+t-1}{2t+1}}・\frac{2t^2+2t-2}{(2t+1)^2}dt\)
\(\displaystyle =\int\frac{2}{t^2+1}dt\)
\(\displaystyle =2\tan^{-1}t+C\)
\(\displaystyle =2\tan^{-1}(\sqrt{x^2+x-1}+x)+C\)
(9)\(\displaystyle \int\frac{1}{(x+1)\sqrt{2+x-x^2}}dx\)
\(\displaystyle t=\sqrt{\frac{2-x}{x+1}}\)とおくと、\(\displaystyle dx=-\frac{6t}{(1+t^2)^2}dt\)
\(\displaystyle =\int\frac{1}{(\frac{3}{1+t^2})^2t}・\frac{-6t}{(1+t^2)^2}dt\)
\(\displaystyle =-\frac{2}{3}\int dt\)
\(\displaystyle =-\frac{2}{3}t+C\)
\(\displaystyle =-\frac{2}{3}\sqrt{\frac{2-x}{x+1}}+C\)
\(\displaystyle =\int\frac{1}{(\frac{3}{1+t^2})^2t}・\frac{-6t}{(1+t^2)^2}dt\)
\(\displaystyle =-\frac{2}{3}\int dt\)
\(\displaystyle =-\frac{2}{3}t+C\)
\(\displaystyle =-\frac{2}{3}\sqrt{\frac{2-x}{x+1}}+C\)
(10)\(\displaystyle \int\frac{1}{x+1}\sqrt{\frac{x}{1-x}}dx\)
\(\displaystyle t=\sqrt{\frac{x}{1-x}}\)とおくと、\(\displaystyle dx=\frac{2t}{(1+t^2)^2}dt\)
\(\displaystyle =\int\frac{1}{\frac{t^2}{1+t^2}+1}・t・\frac{2t}{(1+t^2)^2}dt\)
\(\displaystyle =\int\frac{2t^2}{(1+t^2)(1+2t^2)}dt\)
\(\displaystyle =\int\left(\frac{2}{t^2+1}-\frac{1}{t^2+\frac{1}{2}}\right)dt\)
\(\displaystyle =2\tan^{-1}t-\sqrt{2}\tan^{-1}\sqrt{2}t+C\)
\(\displaystyle =2\tan^{-1}\sqrt{\frac{x}{1-x}}-\sqrt{2}\tan^{-1}\sqrt{\frac{2x}{1-x}}+C\)
\(\displaystyle =\int\frac{1}{\frac{t^2}{1+t^2}+1}・t・\frac{2t}{(1+t^2)^2}dt\)
\(\displaystyle =\int\frac{2t^2}{(1+t^2)(1+2t^2)}dt\)
\(\displaystyle =\int\left(\frac{2}{t^2+1}-\frac{1}{t^2+\frac{1}{2}}\right)dt\)
\(\displaystyle =2\tan^{-1}t-\sqrt{2}\tan^{-1}\sqrt{2}t+C\)
\(\displaystyle =2\tan^{-1}\sqrt{\frac{x}{1-x}}-\sqrt{2}\tan^{-1}\sqrt{\frac{2x}{1-x}}+C\)
(11)\(\displaystyle \int\frac{x^2}{\sqrt{2-x^2}}dx\)
\(\displaystyle x=\sqrt{2}\sin\theta\)とおくと、\(dx=\sqrt{2}\cos\theta d\theta\)
\(\displaystyle =\int\frac{2\sin^2\theta}{\sqrt{2}\cos\theta}・\sqrt{2}\cos\theta d\theta\)
\(\displaystyle =\int2\sin^2\theta d\theta\)
\(\displaystyle =\int(1-\cos2\theta)d\theta\)
\(\displaystyle =\theta-\frac{1}{2}\sin2\theta+C\)
\(\displaystyle =\sin^{-1}\frac{x}{\sqrt{2}}-\frac{1}{2}x\sqrt{2-x^2}+C\)
\(\displaystyle =\int\frac{2\sin^2\theta}{\sqrt{2}\cos\theta}・\sqrt{2}\cos\theta d\theta\)
\(\displaystyle =\int2\sin^2\theta d\theta\)
\(\displaystyle =\int(1-\cos2\theta)d\theta\)
\(\displaystyle =\theta-\frac{1}{2}\sin2\theta+C\)
\(\displaystyle =\sin^{-1}\frac{x}{\sqrt{2}}-\frac{1}{2}x\sqrt{2-x^2}+C\)
(12)\(\displaystyle \int\frac{1}{\sqrt{(x^2+3)^5}}dx\)
\(\displaystyle x=\sqrt{3}\tan\theta\)とおくと、\(\displaystyle dx=\frac{\sqrt{3}}{\cos^2\theta} d\theta\)
\(\displaystyle =\int\frac{1}{\sqrt{(3\tan^2\theta+3)^5}}・\frac{\sqrt{3}}{\cos^2\theta}d\theta\)
\(\displaystyle =\int\frac{\cos^3\theta}{9}d\theta\)
\(\displaystyle =\int\frac{(1-\sin^2\theta)\cos\theta}{9}d\theta\)
\(\displaystyle =\frac{\sin\theta}{9}-\frac{\sin^3\theta}{27}+C\)
\(\displaystyle =\frac{x}{9\sqrt{x^2+3}}-\frac{x^3}{27\sqrt{(x^2+3)^3}}+C\)
\(\displaystyle =\int\frac{1}{\sqrt{(3\tan^2\theta+3)^5}}・\frac{\sqrt{3}}{\cos^2\theta}d\theta\)
\(\displaystyle =\int\frac{\cos^3\theta}{9}d\theta\)
\(\displaystyle =\int\frac{(1-\sin^2\theta)\cos\theta}{9}d\theta\)
\(\displaystyle =\frac{\sin\theta}{9}-\frac{\sin^3\theta}{27}+C\)
\(\displaystyle =\frac{x}{9\sqrt{x^2+3}}-\frac{x^3}{27\sqrt{(x^2+3)^3}}+C\)
(13)\(\displaystyle \int\sqrt{x^2+1}dx\)
\(\displaystyle =x\sqrt{x^2+1}-\int x・\frac{2x}{2\sqrt{x^2+1}}dx\)
\(\displaystyle =x\sqrt{x^2+1}-\int\left(\sqrt{x^2+1}-\frac{1}{\sqrt{x^2+1}}\right)dx\)
\(\displaystyle I=\int\sqrt{x^2+1}dx\)とおくと
\(\displaystyle I=x\sqrt{x^2+1}-I+\log(x+\sqrt{x^2+1})+C\)
\(\displaystyle I=\frac{1}{2}\{x\sqrt{x^2+1}+\log(x+\sqrt{x^2+1})\}+C\)
よって、
\(\displaystyle \int\sqrt{x^2+1}dx\)
\(\displaystyle =\frac{1}{2}\{x\sqrt{x^2+1}+\log(x+\sqrt{x^2+1})\}+C\)
\(\displaystyle =x\sqrt{x^2+1}-\int\left(\sqrt{x^2+1}-\frac{1}{\sqrt{x^2+1}}\right)dx\)
\(\displaystyle I=\int\sqrt{x^2+1}dx\)とおくと
\(\displaystyle I=x\sqrt{x^2+1}-I+\log(x+\sqrt{x^2+1})+C\)
\(\displaystyle I=\frac{1}{2}\{x\sqrt{x^2+1}+\log(x+\sqrt{x^2+1})\}+C\)
よって、
\(\displaystyle \int\sqrt{x^2+1}dx\)
\(\displaystyle =\frac{1}{2}\{x\sqrt{x^2+1}+\log(x+\sqrt{x^2+1})\}+C\)
次の学習に進もう！