与えられた12個の関数をそれぞれ積分する。

1. 問題の内容

与えられた12個の関数をそれぞれ積分する。

2. 解き方の手順

(1) $(x-3)(2x-1) = 2x^2 - 7x + 3$

$\int (2x^2 - 7x + 3) dx = \frac{2}{3}x^3 - \frac{7}{2}x^2 + 3x + C$

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

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

(3) $\frac{(x-1)\sqrt{x}-2}{x^2}$

この関数は積分が難しい。まず、積分記号内に現れる$\sqrt{x}$を$t$と置換して、$x=t^2$なので、$dx = 2t dt$。すると、積分は次のようになります。

$\int \frac{(t^2-1)t-2}{t^4} 2t dt = 2 \int \frac{t^3 - t - 2}{t^3} dt = 2 \int (1 - \frac{1}{t^2} - \frac{2}{t^3}) dt = 2(t + \frac{1}{t} + \frac{1}{t^2}) + C = 2\sqrt{x} + \frac{2}{\sqrt{x}} + \frac{1}{x} + C$.

(4) $(e^x + e^{-x})^2 = e^{2x} + 2 + e^{-2x}$

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

(5) $2^{x+1} + 3^{x+1} = 2 \cdot 2^x + 3 \cdot 3^x$

$\int (2 \cdot 2^x + 3 \cdot 3^x) dx = \frac{2}{\ln 2} 2^x + \frac{3}{\ln 3} 3^x + C$

(6) $\frac{1}{\cos^2 x \sin^2 x} = \frac{\sin^2 x + \cos^2 x}{\cos^2 x \sin^2 x} = \frac{1}{\cos^2 x} + \frac{1}{\sin^2 x} = \sec^2 x + \csc^2 x$

$\int (\sec^2 x + \csc^2 x) dx = \tan x - \cot x + C$

(7) $\frac{\cos^2 x}{2}$

$\int \frac{\cos^2 x}{2} dx = \frac{1}{2} \int \cos^2 x dx = \frac{1}{2} \int \frac{1 + \cos 2x}{2} dx = \frac{1}{4} \int (1 + \cos 2x) dx = \frac{1}{4} (x + \frac{1}{2} \sin 2x) + C = \frac{1}{4}x + \frac{1}{8} \sin 2x + C$

(8) $\sin^3 x = \sin x (1 - \cos^2 x) = \sin x - \sin x \cos^2 x$

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

(9) $\sin x \cos 2x = \sin x (\cos^2 x - \sin^2 x) = \sin x \cos^2 x - \sin^3 x = \sin x \cos^2 x - \sin x (1 - \cos^2 x) = 2 \sin x \cos^2 x - \sin x$

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

(10) $\frac{1}{\tan^2 x} = \cot^2 x = \csc^2 x - 1$

$\int \cot^2 x dx = \int (\csc^2 x - 1) dx = -\cot x - x + C$

(11) $\frac{x + \sqrt{x^2 - 1}}{x - \sqrt{x^2 - 1}} = \frac{(x + \sqrt{x^2 - 1})^2}{x^2 - (x^2 - 1)} = x^2 + 2x \sqrt{x^2 - 1} + x^2 - 1 = 2x^2 - 1 + 2x \sqrt{x^2 - 1}$

$\int \frac{x + \sqrt{x^2 - 1}}{x - \sqrt{x^2 - 1}} dx = \int (2x^2 - 1 + 2x \sqrt{x^2 - 1}) dx = \frac{2}{3}x^3 - x + \frac{2}{3}(x^2 - 1)^{3/2} + C$

(12) $\frac{\sqrt{1 + x^2} + \sqrt{1 - x^2}}{\sqrt{1 - x^4}}$

$\int \frac{\sqrt{1 + x^2} + \sqrt{1 - x^2}}{\sqrt{1 - x^4}} dx = \int \frac{\sqrt{1 + x^2} + \sqrt{1 - x^2}}{\sqrt{(1 - x^2)(1 + x^2)}} dx = \int \frac{\sqrt{1 + x^2}}{\sqrt{(1 - x^2)(1 + x^2)}} dx + \int \frac{\sqrt{1 - x^2}}{\sqrt{(1 - x^2)(1 + x^2)}} dx = \int \frac{1}{\sqrt{1 - x^2}} dx + \int \frac{1}{\sqrt{1 + x^2}} dx = \arcsin x + \text{arcsinh } x + C = \arcsin x + \ln(x + \sqrt{x^2 + 1}) + C$

3. 最終的な答え

(1) $\frac{2}{3}x^3 - \frac{7}{2}x^2 + 3x + C$

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

(3) $2\sqrt{x} + \frac{2}{\sqrt{x}} + \frac{1}{x} + C$

(4) $\frac{1}{2}e^{2x} + 2x - \frac{1}{2}e^{-2x} + C$

(5) $\frac{2}{\ln 2} 2^x + \frac{3}{\ln 3} 3^x + C$

(6) $\tan x - \cot x + C$

(7) $\frac{1}{4}x + \frac{1}{8} \sin 2x + C$

(8) $-\cos x + \frac{1}{3} \cos^3 x + C$

(9) $-\frac{2}{3} \cos^3 x + \cos x + C$

(10) $-\cot x - x + C$

(11) $\frac{2}{3}x^3 - x + \frac{2}{3}(x^2 - 1)^{3/2} + C$

(12) $\arcsin x + \ln(x + \sqrt{x^2 + 1}) + C$