与えられた9つの積分を計算します。 (1) $\int (x^2 + 1) dx$ (2) $\int \cos x dx$ (3) $\int x e^x dx$ (4) $\int 2\sqrt{x} dx$ (5) $\int \frac{1}{x \log x} dx$ (6) $\int \sqrt{2x-3} dx$ (7) $\int_{-1}^{2} 4x dx$ (8) $\int_{0}^{\pi} \sin x dx$ (9) $\int_{1}^{e} \frac{(\log x)^2}{x} dx$

1. 問題の内容

与えられた9つの積分を計算します。

(1) $\int (x^2 + 1) dx$

(2) $\int \cos x dx$

(3) $\int x e^x dx$

(4) $\int 2\sqrt{x} dx$

(5) $\int \frac{1}{x \log x} dx$

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

(7) $\int_{-1}^{2} 4x dx$

(8) $\int_{0}^{\pi} \sin x dx$

(9) $\int_{1}^{e} \frac{(\log x)^2}{x} dx$

2. 解き方の手順

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

(2) $\int \cos x dx = \sin x + C$

(3) $\int x e^x dx$ （部分積分）

$u = x, dv = e^x dx$ とすると、$du = dx, v = e^x$ なので、

$\int x e^x dx = x e^x - \int e^x dx = x e^x - e^x + C$

(4) $\int 2\sqrt{x} dx = 2 \int x^{1/2} dx = 2 \cdot \frac{x^{3/2}}{3/2} + C = \frac{4}{3} x^{3/2} + C$

(5) $\int \frac{1}{x \log x} dx$

$u = \log x$ とすると、$du = \frac{1}{x} dx$ なので、

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

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

$u = 2x-3$ とすると、$du = 2 dx$ より、$dx = \frac{1}{2} du$ なので、

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

(7) $\int_{-1}^{2} 4x dx = [2x^2]_{-1}^{2} = 2(2^2) - 2(-1)^2 = 2(4) - 2(1) = 8 - 2 = 6$

(8) $\int_{0}^{\pi} \sin x dx = [-\cos x]_{0}^{\pi} = -\cos \pi - (-\cos 0) = -(-1) - (-1) = 1 + 1 = 2$

(9) $\int_{1}^{e} \frac{(\log x)^2}{x} dx$

$u = \log x$ とすると、$du = \frac{1}{x} dx$

$x = 1$ のとき $u = \log 1 = 0$

$x = e$ のとき $u = \log e = 1$

$\int_{1}^{e} \frac{(\log x)^2}{x} dx = \int_{0}^{1} u^2 du = [\frac{u^3}{3}]_{0}^{1} = \frac{1^3}{3} - \frac{0^3}{3} = \frac{1}{3}$

3. 最終的な答え

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

(2) $\sin x + C$

(3) $x e^x - e^x + C$

(4) $\frac{4}{3} x^{3/2} + C$

(5) $\log |\log x| + C$

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

(7) $6$

(8) $2$

(9) $\frac{1}{3}$