与えられた18個の関数の不定積分を求める問題です。

1. 問題の内容

与えられた18個の関数の不定積分を求める問題です。

2. 解き方の手順

各関数の不定積分を求めます。積分定数は省略します。

(1) $\int x(2x+1)^8 dx$

$t = 2x+1$ と置換すると、$x = \frac{t-1}{2}$、$dx = \frac{dt}{2}$ より、

$\int \frac{t-1}{2} t^8 \frac{dt}{2} = \frac{1}{4} \int (t^9 - t^8) dt = \frac{1}{4} (\frac{t^{10}}{10} - \frac{t^9}{9}) = \frac{1}{4} (\frac{(2x+1)^{10}}{10} - \frac{(2x+1)^9}{9}) = \frac{(2x+1)^{10}}{40} - \frac{(2x+1)^9}{36}$

(2) $\int \sin(3x+1) dx = -\frac{1}{3} \cos(3x+1)$

(3) $\int \frac{(\log x)^2}{x} dx$

$t = \log x$ と置換すると、$dt = \frac{1}{x} dx$ より、

$\int t^2 dt = \frac{t^3}{3} = \frac{(\log x)^3}{3}$

(4) $\int \frac{e^x}{1+e^x} dx$

$t = 1+e^x$ と置換すると、$dt = e^x dx$ より、

$\int \frac{1}{t} dt = \log |t| = \log (1+e^x)$

(5) $\int x \log x dx = \frac{x^2}{2} \log x - \int \frac{x^2}{2} \frac{1}{x} dx = \frac{x^2}{2} \log x - \int \frac{x}{2} dx = \frac{x^2}{2} \log x - \frac{x^2}{4}$

(6) $\int x \tan^{-1} x dx = \frac{x^2}{2} \tan^{-1} x - \int \frac{x^2}{2} \frac{1}{1+x^2} dx = \frac{x^2}{2} \tan^{-1} x - \frac{1}{2} \int \frac{x^2+1-1}{x^2+1} dx = \frac{x^2}{2} \tan^{-1} x - \frac{1}{2} \int (1 - \frac{1}{x^2+1}) dx = \frac{x^2}{2} \tan^{-1} x - \frac{1}{2} (x - \tan^{-1} x) = \frac{x^2+1}{2} \tan^{-1} x - \frac{x}{2}$

(7) $\int \frac{1}{x^2-2x-3} dx = \int \frac{1}{(x-3)(x+1)} dx = \int \frac{1/4}{x-3} - \frac{1/4}{x+1} dx = \frac{1}{4} \log |x-3| - \frac{1}{4} \log |x+1| = \frac{1}{4} \log |\frac{x-3}{x+1}|$

(8) $\int \frac{1}{x^4+1} dx$

(9) $\int \frac{1}{x^2-1} dx = \int \frac{1/2}{x-1} - \frac{1/2}{x+1} dx = \frac{1}{2} \log |x-1| - \frac{1}{2} \log |x+1| = \frac{1}{2} \log |\frac{x-1}{x+1}|$

(10) $\int \frac{1}{(x^2+1)(x^2+4)} dx = \int \frac{1/3}{x^2+1} - \frac{1/3}{x^2+4} dx = \frac{1}{3} \tan^{-1} x - \frac{1}{6} \tan^{-1} \frac{x}{2}$

(11) $\int \frac{x+3}{x^2+x+4} dx = \frac{1}{2} \int \frac{2x+1+5}{x^2+x+4} dx = \frac{1}{2} \log (x^2+x+4) + \frac{5}{2} \int \frac{1}{x^2+x+4} dx = \frac{1}{2} \log (x^2+x+4) + \frac{5}{2} \int \frac{1}{(x+1/2)^2 + 15/4} dx = \frac{1}{2} \log (x^2+x+4) + \frac{5}{2} \frac{2}{\sqrt{15}} \tan^{-1} \frac{2x+1}{\sqrt{15}} = \frac{1}{2} \log (x^2+x+4) + \frac{5}{\sqrt{15}} \tan^{-1} \frac{2x+1}{\sqrt{15}}$

(12) $\int \frac{\sin x}{2+\tan^2 x} dx = \int \frac{\sin x}{2 + \frac{\sin^2 x}{\cos^2 x}} dx = \int \frac{\sin x \cos^2 x}{2\cos^2 x + \sin^2 x} dx = \int \frac{\sin x \cos^2 x}{2\cos^2 x + (1-\cos^2 x)} dx = \int \frac{\sin x \cos^2 x}{\cos^2 x + 1} dx$

$t = \cos x$ と置換すると、$dt = -\sin x dx$ より、

$\int \frac{-t^2}{t^2+1} dt = -\int (1 - \frac{1}{t^2+1}) dt = -(t - \tan^{-1} t) = -(\cos x - \tan^{-1} (\cos x)) = \tan^{-1} (\cos x) - \cos x$

(13) $\int \frac{\cos^3 x}{\sin^2 x} dx = \int \frac{\cos x (1-\sin^2 x)}{\sin^2 x} dx = \int \frac{\cos x}{\sin^2 x} - \cos x dx$

$t = \sin x$ と置換すると、$dt = \cos x dx$ より、

$\int \frac{1}{t^2} - dt = -\frac{1}{t} - \sin x = -\frac{1}{\sin x} - \sin x$

(14) $\int \sin (\log x) dx = x \sin (\log x) - \int x \cos (\log x) \frac{1}{x} dx = x \sin (\log x) - \int \cos (\log x) dx = x \sin (\log x) - (x \cos (\log x) + \int x (-\sin (\log x)) \frac{1}{x} dx) = x \sin (\log x) - x \cos (\log x) - \int \sin (\log x) dx$

$2 \int \sin (\log x) dx = x \sin (\log x) - x \cos (\log x)$

$\int \sin (\log x) dx = \frac{x}{2} (\sin (\log x) - \cos (\log x))$

(15) $\int \tan^{-1} \sqrt{x} dx = x \tan^{-1} \sqrt{x} - \int \frac{x}{1+x} \frac{1}{2\sqrt{x}} dx$

(16) $\int \frac{1}{x \sqrt{1+x^2}} dx$

$t = \frac{1}{x}$ と置換すると、$x = \frac{1}{t}$、$dx = -\frac{dt}{t^2}$ より、

$\int \frac{1}{\frac{1}{t} \sqrt{1+\frac{1}{t^2}}} (-\frac{dt}{t^2}) = \int \frac{1}{\frac{1}{t} \frac{\sqrt{t^2+1}}{t}} (-\frac{dt}{t^2}) = -\int \frac{1}{\frac{\sqrt{t^2+1}}{t^2}} \frac{dt}{t^2} = -\int \frac{dt}{\sqrt{t^2+1}} = -\sinh^{-1} t = -\sinh^{-1} \frac{1}{x} = -\log (\frac{1}{x} + \sqrt{\frac{1}{x^2}+1}) = -\log (\frac{1+\sqrt{1+x^2}}{x}) = \log \frac{x}{1+\sqrt{1+x^2}}$

(17) $\int \frac{\sqrt{1+\log x}}{x} dx$

$t = 1 + \log x$ と置換すると、$dt = \frac{1}{x} dx$ より、

$\int \sqrt{t} dt = \frac{2}{3} t^{3/2} = \frac{2}{3} (1+\log x)^{3/2}$

(18) $\int \sqrt{4x-x^2} dx = \int \sqrt{4 - (x-2)^2} dx$

$x-2 = 2\sin \theta$ と置換すると、$dx = 2\cos \theta d\theta$ より、

$\int \sqrt{4 - 4\sin^2 \theta} 2\cos \theta d\theta = \int 2\cos \theta 2\cos \theta d\theta = 4\int \cos^2 \theta d\theta = 4\int \frac{1+\cos 2\theta}{2} d\theta = 2 (\theta + \frac{\sin 2\theta}{2}) = 2\theta + \sin 2\theta = 2\theta + 2\sin \theta \cos \theta = 2\theta + 2\sin \theta \sqrt{1-\sin^2 \theta} = 2\arcsin \frac{x-2}{2} + 2 \frac{x-2}{2} \sqrt{1 - (\frac{x-2}{2})^2} = 2\arcsin \frac{x-2}{2} + (x-2) \frac{\sqrt{4-(x-2)^2}}{2} = 2\arcsin \frac{x-2}{2} + \frac{(x-2)}{2} \sqrt{4x-x^2}$

3. 最終的な答え

(1) $\frac{(2x+1)^{10}}{40} - \frac{(2x+1)^9}{36}$

(2) $-\frac{1}{3} \cos(3x+1)$

(3) $\frac{(\log x)^3}{3}$

(4) $\log (1+e^x)$

(5) $\frac{x^2}{2} \log x - \frac{x^2}{4}$

(6) $\frac{x^2+1}{2} \tan^{-1} x - \frac{x}{2}$

(7) $\frac{1}{4} \log |\frac{x-3}{x+1}|$

(8) $\int \frac{1}{x^4+1} dx$

(9) $\frac{1}{2} \log |\frac{x-1}{x+1}|$

(10) $\frac{1}{3} \tan^{-1} x - \frac{1}{6} \tan^{-1} \frac{x}{2}$

(11) $\frac{1}{2} \log (x^2+x+4) + \frac{5}{\sqrt{15}} \tan^{-1} \frac{2x+1}{\sqrt{15}}$

(12) $\tan^{-1} (\cos x) - \cos x$

(13) $-\frac{1}{\sin x} - \sin x$

(14) $\frac{x}{2} (\sin (\log x) - \cos (\log x))$

(15) $x \tan^{-1} \sqrt{x} - \int \frac{x}{1+x} \frac{1}{2\sqrt{x}} dx$

(16) $\log \frac{x}{1+\sqrt{1+x^2}}$

(17) $\frac{2}{3} (1+\log x)^{3/2}$

(18) $2\arcsin \frac{x-2}{2} + \frac{(x-2)}{2} \sqrt{4x-x^2}$