第8章 8.1

(1)

∫(6x2−1)dx=6⋅13x3−x+C=2x3−x+C\int (6x^2-1)dx = 6\cdot\frac{1}{3}x^3 - x + C = 2x^3 - x + C

(2)

∫(x3−5x)dx=14x4−5⋅12x2+C=14x4−52x2+C\int (x^3-5x)dx = \frac{1}{4}x^4 - 5\cdot\frac{1}{2}x^2 + C = \frac{1}{4}x^4 - \frac{5}{2}x^2 + C

(3)

∫(2−x)(x−1)dx=∫(−x2+3x−2)dx=−13x3+3⋅12x2−2x+C=−13x3+32x2−2x+C\int (2-x)(x-1)dx = \int (-x^2+3x-2)dx = -\frac{1}{3}x^3 + 3\cdot\frac{1}{2}x^2 - 2x + C = -\frac{1}{3}x^3 + \frac{3}{2}x^2 - 2x + C

(4)

∫(x−1x2)dx=12x2−(−1)1x+C=12x2+1x+C\int \left(x - \frac{1}{x^2}\right)dx = \frac{1}{2}x^2 - (-1)\frac{1}{x} + C = \frac{1}{2}x^2 + \frac{1}{x} + C

(5)

∫(2x−x)dx=2⋅12x2−23x3+C=x2−23x3+C\int (2x - \sqrt{x})dx = 2\cdot\frac{1}{2}x^2 - \frac{2}{3}\sqrt{x^3} + C = x^2 - \frac{2}{3}\sqrt{x^3} + C

(6)

∫(x−1x)2dx=∫(x2−2x+1x)dx=13x3−43x3+log⁡x+C\int \left(x - \frac{1}{\sqrt{x}}\right)^2 dx = \int \left(x^2 - 2\sqrt{x} + \frac{1}{x}\right)dx = \frac{1}{3}x^3 - \frac{4}{3}\sqrt{x^3} + \log x + C

解説: r31bn1z