次の6つの関数の不定積分を求める問題です。 (1) $f(x) = \frac{2x-3}{x^3+x^2-2}$ (2) $f(x) = e^{3x} \sin x$ (3) $f(x) = x e^{x^2}$ (4) $f(x) = \frac{1}{(1-x)\sqrt{1+x}}$ (5) $f(x) = x^2 e^x$ (6) $f(x) = \frac{x^4}{x^2+1}$

1. 問題の内容

次の6つの関数の不定積分を求める問題です。

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

(2) $f(x) = e^{3x} \sin x$

(3) $f(x) = x e^{x^2}$

(4) $f(x) = \frac{1}{(1-x)\sqrt{1+x}}$

(5) $f(x) = x^2 e^x$

(6) $f(x) = \frac{x^4}{x^2+1}$

2. 解き方の手順

(1) 部分分数分解を用いる。$x^3+x^2-2 = (x-1)(x^2+2x+2)$ より、

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

とおき、A, B, C を求める。

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

係数を比較して、

$A+B = 0$, $2A-B+C=2$, $2A-C = -3$

この連立方程式を解くと、$A=-1/5$, $B=1/5$, $C=-13/5$ となる。

よって、

$\int \frac{2x-3}{x^3+x^2-2} dx = \int (-\frac{1}{5(x-1)} + \frac{x-13}{5(x^2+2x+2)}) dx = -\frac{1}{5} \int \frac{1}{x-1} dx + \frac{1}{5} \int \frac{x-13}{x^2+2x+2} dx$

$x^2+2x+2 = (x+1)^2 + 1$ より、$x+1=u$ とおくと、$x = u-1$, $dx = du$

$\frac{x-13}{x^2+2x+2} = \frac{u-14}{u^2+1}$

$\int \frac{u-14}{u^2+1} du = \int \frac{u}{u^2+1} du - 14 \int \frac{1}{u^2+1} du = \frac{1}{2} \log(u^2+1) - 14 \arctan(u)$

$=\frac{1}{2} \log((x+1)^2+1) - 14 \arctan(x+1) = \frac{1}{2} \log(x^2+2x+2) - 14 \arctan(x+1)$

最終的に、

$\int \frac{2x-3}{x^3+x^2-2} dx = -\frac{1}{5} \log|x-1| + \frac{1}{10} \log(x^2+2x+2) - \frac{14}{5} \arctan(x+1) + C$

(2) 部分積分を2回行う。$I = \int e^{3x} \sin x dx$ とおく。

$I = \int e^{3x} \sin x dx = e^{3x}(-\cos x) - \int 3e^{3x} (-\cos x) dx = -e^{3x}\cos x + 3 \int e^{3x} \cos x dx$

$= -e^{3x}\cos x + 3 (e^{3x}\sin x - \int 3e^{3x} \sin x dx) = -e^{3x}\cos x + 3 e^{3x}\sin x - 9 \int e^{3x} \sin x dx = -e^{3x}\cos x + 3 e^{3x}\sin x - 9I$

$10I = e^{3x}(3\sin x - \cos x)$

$I = \frac{e^{3x}(3\sin x - \cos x)}{10} + C$

(3) 置換積分を行う。$u = x^2$ とおくと、$du = 2x dx$, $xdx = \frac{1}{2}du$

$\int xe^{x^2} dx = \int e^u \frac{1}{2} du = \frac{1}{2} e^u + C = \frac{1}{2} e^{x^2} + C$

(4) 置換積分を行う。$t = \sqrt{1+x}$ とおくと、$t^2 = 1+x$, $x = t^2 - 1$, $dx = 2t dt$, $1-x = 1-(t^2-1) = 2-t^2$

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

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

$= \frac{1}{\sqrt{2}} \log|\frac{\sqrt{2}+\sqrt{1+x}}{\sqrt{2}-\sqrt{1+x}}| + C$

(5) 部分積分を2回行う。

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

(6) 割り算を行う。

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

$\int \frac{x^4}{x^2+1} dx = \int (x^2-1 + \frac{1}{x^2+1}) dx = \frac{x^3}{3} - x + \arctan(x) + C$

3. 最終的な答え

(1) $\int \frac{2x-3}{x^3+x^2-2} dx = -\frac{1}{5} \log|x-1| + \frac{1}{10} \log(x^2+2x+2) - \frac{14}{5} \arctan(x+1) + C$

(2) $\int e^{3x} \sin x dx = \frac{e^{3x}(3\sin x - \cos x)}{10} + C$

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

(4) $\int \frac{1}{(1-x)\sqrt{1+x}} dx = \frac{1}{\sqrt{2}} \log|\frac{\sqrt{2}+\sqrt{1+x}}{\sqrt{2}-\sqrt{1+x}}| + C$

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

(6) $\int \frac{x^4}{x^2+1} dx = \frac{x^3}{3} - x + \arctan(x) + C$