次の不定積分を求めよ。 (1) $\int (5x^4 + \frac{4}{x} - \frac{1}{x^4} - 2) dx$ (2) $\int (4x^3 + 3x^{1/2} - 1) dx$ (3) $\int \frac{\sqrt{x} - \frac{1}{\sqrt{x}}}{2} dx$

1. 問題の内容

次の不定積分を求めよ。

(1) $\int (5x^4 + \frac{4}{x} - \frac{1}{x^4} - 2) dx$

(2) $\int (4x^3 + 3x^{1/2} - 1) dx$

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

2. 解き方の手順

(1)

$\int (5x^4 + \frac{4}{x} - \frac{1}{x^4} - 2) dx = \int (5x^4 + 4x^{-1} - x^{-4} - 2) dx$

$= 5\int x^4 dx + 4\int \frac{1}{x} dx - \int x^{-4} dx - 2\int dx$

$= 5 \cdot \frac{x^5}{5} + 4 \ln|x| - \frac{x^{-3}}{-3} - 2x + C$

$= x^5 + 4 \ln|x| + \frac{1}{3x^3} - 2x + C$

(2)

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

$= 4 \cdot \frac{x^4}{4} + 3 \cdot \frac{x^{3/2}}{3/2} - x + C$

$= x^4 + 2x^{3/2} - x + C$

(3)

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

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

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

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

$= \frac{1}{3}x^{3/2} - x^{1/2} + C$

$= \frac{1}{3}x\sqrt{x} - \sqrt{x} + C$

3. 最終的な答え

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

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

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