与えられた不定積分の中から、以下の問題を解きます。 (1) $\int \frac{x+1}{x^2+x-6} dx$ (2) $\int \sin 2x \cos 3x dx$ (3) $\int \frac{\sqrt{x+2} - \sqrt{2}}{x} dx$ (4) $\int \frac{x^2}{\sqrt{x-1}} dx$ (5) $\int \frac{\sin x}{\cos^3 x} dx$ (6) $\int 2x^2 e^x dx$

1. 問題の内容

与えられた不定積分の中から、以下の問題を解きます。

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

(2) $\int \sin 2x \cos 3x dx$

(3) $\int \frac{\sqrt{x+2} - \sqrt{2}}{x} dx$

(4) $\int \frac{x^2}{\sqrt{x-1}} dx$

(5) $\int \frac{\sin x}{\cos^3 x} dx$

(6) $\int 2x^2 e^x dx$

2. 解き方の手順

(1) 部分分数分解を行います。

$x^2 + x - 6 = (x+3)(x-2)$なので、

$\frac{x+1}{x^2+x-6} = \frac{A}{x+3} + \frac{B}{x-2}$とおく。

$x+1 = A(x-2) + B(x+3)$

$x=2$のとき、$3 = 5B$より$B = \frac{3}{5}$

$x=-3$のとき、$-2 = -5A$より$A = \frac{2}{5}$

$\int \frac{x+1}{x^2+x-6} dx = \int (\frac{2/5}{x+3} + \frac{3/5}{x-2}) dx$

$= \frac{2}{5} \int \frac{1}{x+3} dx + \frac{3}{5} \int \frac{1}{x-2} dx$

$= \frac{2}{5} \ln|x+3| + \frac{3}{5} \ln|x-2| + C$

(2) 積を和の形に直します。

$\sin 2x \cos 3x = \frac{1}{2} (\sin (2x+3x) + \sin (2x-3x)) = \frac{1}{2} (\sin 5x - \sin x)$

$\int \sin 2x \cos 3x dx = \frac{1}{2} \int (\sin 5x - \sin x) dx$

$= \frac{1}{2} (-\frac{1}{5} \cos 5x + \cos x) + C$

$= -\frac{1}{10} \cos 5x + \frac{1}{2} \cos x + C$

(3) $t = \sqrt{x+2}$とおくと、$t^2 = x+2$、$x = t^2-2$、$dx = 2t dt$

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

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

$\frac{-\sqrt{2}t+2}{(t-\sqrt{2})(t+\sqrt{2})} = \frac{A}{t-\sqrt{2}} + \frac{B}{t+\sqrt{2}}$とおくと

$-\sqrt{2}t+2 = A(t+\sqrt{2}) + B(t-\sqrt{2})$

$t = \sqrt{2}$のとき、 $0 = 2\sqrt{2}A$より$A = 0$

$t = -\sqrt{2}$のとき、$4 = -2\sqrt{2}B$より$B = -\sqrt{2}$

$= 2 \int dt + 2 \int \frac{-\sqrt{2}}{t+\sqrt{2}} dt = 2t - 2\sqrt{2} \ln |t+\sqrt{2}| + C = 2\sqrt{x+2} - 2\sqrt{2} \ln |\sqrt{x+2} + \sqrt{2}| + C$

(4) $t = \sqrt{x-1}$とおくと、$t^2 = x-1$、$x = t^2 + 1$、$dx = 2t dt$

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

$= 2(\frac{1}{5}t^5 + \frac{2}{3}t^3 + t) + C = \frac{2}{5}(x-1)^{\frac{5}{2}} + \frac{4}{3}(x-1)^{\frac{3}{2}} + 2\sqrt{x-1} + C$

(5) $t = \cos x$とおくと、$dt = -\sin x dx$

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

(6) 部分積分を繰り返します。

$\int 2x^2 e^x dx = 2\int x^2 e^x dx$

$u = x^2, dv = e^x dx$とおくと、$du = 2x dx, v = e^x$

$2(x^2 e^x - \int 2x e^x dx) = 2x^2 e^x - 4\int x e^x dx$

$u = x, dv = e^x dx$とおくと、$du = dx, v = e^x$

$2x^2 e^x - 4(xe^x - \int e^x dx) = 2x^2 e^x - 4xe^x + 4e^x + C = 2e^x(x^2 - 2x + 2) + C$

3. 最終的な答え

(1) $\frac{2}{5} \ln|x+3| + \frac{3}{5} \ln|x-2| + C$

(2) $-\frac{1}{10} \cos 5x + \frac{1}{2} \cos x + C$

(3) $2\sqrt{x+2} - 2\sqrt{2} \ln |\sqrt{x+2} + \sqrt{2}| + C$

(4) $\frac{2}{5}(x-1)^{\frac{5}{2}} + \frac{4}{3}(x-1)^{\frac{3}{2}} + 2\sqrt{x-1} + C$

(5) $\frac{1}{2\cos^2 x} + C$

(6) $2e^x(x^2 - 2x + 2) + C$