次の計算をせよ。 (1) $\frac{d}{dx}\log_e |2x|$ (2) $\frac{d}{dx}\log_2 x$ (3) $\int x\sqrt{x} dx$ (4) $\int_0^1 3^{2x} dx$

1. 問題の内容

次の計算をせよ。

(1) $\frac{d}{dx}\log_e |2x|$

(2) $\frac{d}{dx}\log_2 x$

(3) $\int x\sqrt{x} dx$

(4) $\int_0^1 3^{2x} dx$

2. 解き方の手順

(1) $\frac{d}{dx}\log_e |2x|$

$\log_e |2x| = \log_e 2 + \log_e |x|$

$\frac{d}{dx}\log_e |2x| = \frac{d}{dx}(\log_e 2 + \log_e |x|) = \frac{d}{dx}\log_e |x|$

$x>0$ のとき $\log_e |x| = \log_e x$ なので $\frac{d}{dx}\log_e x = \frac{1}{x}$

$x<0$ のとき $\log_e |x| = \log_e (-x)$ なので $\frac{d}{dx}\log_e (-x) = \frac{1}{-x}(-1) = \frac{1}{x}$

よって $\frac{d}{dx}\log_e |2x| = \frac{1}{x}$

(2) $\frac{d}{dx}\log_2 x$

$\log_2 x = \frac{\log_e x}{\log_e 2}$

$\frac{d}{dx}\log_2 x = \frac{d}{dx}(\frac{\log_e x}{\log_e 2}) = \frac{1}{\log_e 2}\frac{d}{dx}\log_e x = \frac{1}{\log_e 2} \cdot \frac{1}{x} = \frac{1}{x\log_e 2}$

(3) $\int x\sqrt{x} dx$

$\int x\sqrt{x} dx = \int x^{3/2} dx = \frac{x^{5/2}}{5/2} + C = \frac{2}{5}x^{5/2} + C$

(4) $\int_0^1 3^{2x} dx$

$3^{2x} = (3^2)^x = 9^x$

$\int 9^x dx = \int e^{x\log_e 9} dx = \frac{e^{x\log_e 9}}{\log_e 9} + C = \frac{9^x}{\log_e 9} + C$

$\int_0^1 3^{2x} dx = \int_0^1 9^x dx = [\frac{9^x}{\log_e 9}]_0^1 = \frac{9^1}{\log_e 9} - \frac{9^0}{\log_e 9} = \frac{9}{\log_e 9} - \frac{1}{\log_e 9} = \frac{8}{\log_e 9} = \frac{8}{\log_e 3^2} = \frac{8}{2\log_e 3} = \frac{4}{\log_e 3}$

3. 最終的な答え

(1) $\frac{1}{x}$

(2) $\frac{1}{x\log_e 2}$

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

(4) $\frac{4}{\log_e 3}$