与えられた8つの関数の不定積分を求める問題です。 (1) $x(2x+1)^8$ (2) $\sin(3x+1)$ (3) $\frac{(\log x)^2}{x}$ (4) $\frac{e^x}{1+e^x}$ (5) $x \log x$ (6) $x \tan^{-1} x$ (7) $\frac{1}{x^2-2x-3}$ (8) $\frac{1}{x^4+1}$

1. 問題の内容

与えられた8つの関数の不定積分を求める問題です。

(1) $x(2x+1)^8$

(2) $\sin(3x+1)$

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

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

(5) $x \log x$

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

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

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

2. 解き方の手順

(1) $x(2x+1)^8$

$u = 2x+1$ と置換すると、$x = \frac{u-1}{2}$、$dx = \frac{1}{2}du$ となります。

したがって、

$\int x(2x+1)^8 dx = \int \frac{u-1}{2}u^8 \frac{1}{2} du = \frac{1}{4}\int (u^9 - u^8) du = \frac{1}{4}(\frac{u^{10}}{10} - \frac{u^9}{9}) + C = \frac{1}{4}(\frac{(2x+1)^{10}}{10} - \frac{(2x+1)^9}{9}) + C = \frac{(2x+1)^9}{360} (9(2x+1)-10)+C = \frac{(2x+1)^9(18x-1)}{360} + C$

(2) $\sin(3x+1)$

$u = 3x+1$ と置換すると、$du = 3dx$ より、$dx = \frac{1}{3}du$ となります。

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

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

$u = \log x$ と置換すると、$du = \frac{1}{x}dx$ となります。

$\int \frac{(\log x)^2}{x} dx = \int u^2 du = \frac{u^3}{3} + C = \frac{(\log x)^3}{3} + C$

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

$u = 1+e^x$ と置換すると、$du = e^x dx$ となります。

$\int \frac{e^x}{1+e^x} dx = \int \frac{1}{u} du = \log |u| + C = \log(1+e^x) + C$

(5) $x \log x$

部分積分を行います。$u = \log x, dv = x dx$ とすると、$du = \frac{1}{x}dx, v = \frac{x^2}{2}$ となります。

$\int x \log x dx = \frac{x^2}{2} \log x - \int \frac{x^2}{2} \frac{1}{x} dx = \frac{x^2}{2} \log x - \int \frac{x}{2} dx = \frac{x^2}{2} \log x - \frac{x^2}{4} + C$

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

部分積分を行います。$u = \tan^{-1} x, dv = x dx$ とすると、$du = \frac{1}{1+x^2}dx, v = \frac{x^2}{2}$ となります。

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

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

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

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

$x = 3$ のとき、$1 = 4A$ より $A = \frac{1}{4}$

$x = -1$ のとき、$1 = -4B$ より $B = -\frac{1}{4}$

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

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

$\frac{1}{x^4+1} = \frac{Ax+B}{x^2+\sqrt{2}x+1}+\frac{Cx+D}{x^2-\sqrt{2}x+1}$

$1 = (Ax+B)(x^2-\sqrt{2}x+1)+(Cx+D)(x^2+\sqrt{2}x+1)$

$A+C=0, -\sqrt{2}A+B+\sqrt{2}C+D=0, A-\sqrt{2}B+C+\sqrt{2}D=0, B+D=1$

$A=-C, 2\sqrt{2}C+B+D=0, 2A+\sqrt{2}(D-B)=0$

$B+D=1, 2\sqrt{2}C+1=0$

$C = -\frac{1}{2\sqrt{2}} = -\frac{\sqrt{2}}{4}, A = \frac{\sqrt{2}}{4}$

$2A+\sqrt{2}(D-B)=0$

$2 \cdot \frac{\sqrt{2}}{4} + \sqrt{2}(D-B) = 0$

$\frac{\sqrt{2}}{2} + \sqrt{2}(D-B) = 0$

$D-B = -\frac{1}{2}$

$B+D = 1, D-B = -\frac{1}{2}$ より $2D = \frac{1}{2}, D = \frac{1}{4}$

$B = 1 - D = \frac{3}{4}$

$\int \frac{1}{x^4+1}dx = \frac{\sqrt{2}}{4}\int \frac{x+\frac{3}{\sqrt{2}}}{x^2+\sqrt{2}x+1}dx - \frac{\sqrt{2}}{4}\int \frac{x-\frac{3}{\sqrt{2}}}{x^2-\sqrt{2}x+1}dx$

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

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

$\int \frac{1}{x^4+1}dx= \frac{\sqrt{2}}{8}log(\frac{x^2+\sqrt{2}x+1}{x^2-\sqrt{2}x+1}) + \frac{\sqrt{2}}{4}\cdot (\frac{1}{\sqrt{2}})(\arctan (\sqrt{2} x+1)+\arctan (\sqrt{2} x-1))$

$=\frac{\sqrt{2}}{8}\log(\frac{x^2+\sqrt{2}x+1}{x^2-\sqrt{2}x+1})+\frac{\sqrt{2}}{4}atan \left ( \frac{2x\sqrt{2}}{-2x^2} \right )$

$\frac{\sqrt{2}}{8}\log(\frac{x^2+\sqrt{2}x+1}{x^2-\sqrt{2}x+1})+\frac{\sqrt{2}}{4}\tan^{-1}(\sqrt{2}x^2)+\frac{1}{\sqrt{2}\frac{\pi}{4}}+ C$

3. 最終的な答え

(1) $\frac{(2x+1)^9(18x-1)}{360} + C$

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

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

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

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

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

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

(8) $\frac{\sqrt{2}}{8}\log\left(\frac{x^2+\sqrt{2}x+1}{x^2-\sqrt{2}x+1}\right) + \frac{\sqrt{2}}{4}\arctan\left(\frac{2x}{1-x^2}\right) + C$