高専の数学解説

第4章 4.9

(1)

1(x−a)(x−b)(a≠b)\frac{1}{(x-a)(x-b)} \quad (a \neq b)

∫1(x−a)(x−b)dx=∫1a−b(1x−a−1x−b)dx=1a−blog⁡∣x−ax−b∣+C(C は積分定数)\int \frac{1}{(x-a)(x-b)} dx = \int \frac{1}{a-b}\left(\frac{1}{x-a} - \frac{1}{x-b}\right) dx = \frac{1}{a-b}\log\left|\frac{x-a}{x-b}\right| + C \qquad \text{(C \text{は積分定数})}

(2)

x(x−a)(x−b)(a≠b)\frac{x}{(x-a)(x-b)} \quad (a \neq b)

∫x(x−a)(x−b)dx=∫1a−b(ax−a−bx−b)dx=1a−b(alog⁡∣x−a∣−blog⁡∣x−b∣)+C(C は積分定数)\int \frac{x}{(x-a)(x-b)} dx = \int \frac{1}{a-b}\left(\frac{a}{x-a} - \frac{b}{x-b}\right) dx = \frac{1}{a-b}(a\log|x-a| - b\log|x-b|) + C \qquad \text{(C \text{は積分定数})}

(3)

x2+6x+11(x+2)3\frac{x^2+6x+11}{(x+2)^3}

∫x2+6x+11(x+2)3dx=∫{1x+2+2x+7(x+2)3}dx=∫{1x+2+7(x+2)3+2x(x+2)3}dx\int \frac{x^2+6x+11}{(x+2)^3} dx = \int \left\{\frac{1}{x+2} + \frac{2x+7}{(x+2)^3}\right\} dx = \int \left\{\frac{1}{x+2} + \frac{7}{(x+2)^3} + \frac{2x}{(x+2)^3}\right\} dx

=∫{1x+2+7(x+2)3+(2(x+2)2+−4(x+2)3)}dx= \int \left\{\frac{1}{x+2} + \frac{7}{(x+2)^3} + \left(\frac{2}{(x+2)^2} + \frac{-4}{(x+2)^3}\right)\right\} dx

=log⁡∣x+2∣−72⋅1(x+2)2−2x+2+2(x+2)2+C=log⁡∣x+2∣−2x+2−32(x+2)2+C= \log|x+2| - \frac{7}{2}\cdot\frac{1}{(x+2)^2} - \frac{2}{x+2} + \frac{2}{(x+2)^2} + C = \log|x+2| - \frac{2}{x+2} - \frac{3}{2(x+2)^2} + C

(C は積分定数)

(4)

5x−1x3−3x−2\frac{5x-1}{x^3-3x-2}

∫5x−1x3−3x−2dx=∫{1x−2+−x+1(x+1)2}dx=∫{1x−2−12⋅2x+2(x+1)2+2(x+1)2}dx\int \frac{5x-1}{x^3-3x-2} dx = \int \left\{\frac{1}{x-2} + \frac{-x+1}{(x+1)^2}\right\} dx = \int \left\{\frac{1}{x-2} - \frac{1}{2}\cdot\frac{2x+2}{(x+1)^2} + \frac{2}{(x+1)^2}\right\} dx

=log⁡∣x−2∣−12log⁡(x+1)2−2x+1+C=log⁡∣x−2x+1∣−2x+1+C(C は積分定数)= \log|x-2| - \frac{1}{2}\log(x+1)^2 - \frac{2}{x+1} + C = \log\left|\frac{x-2}{x+1}\right| - \frac{2}{x+1} + C \qquad \text{(C \text{は積分定数})}

(5)

5x+2x2+2x+10\frac{5x+2}{x^2+2x+10}

∫5x+2x2+2x+10dx=∫{52⋅2x+2x2+2x+10−3x2+2x+10}dx=∫{52⋅2x+2x2+2x+10−3(x+1)2+9}dx\int \frac{5x+2}{x^2+2x+10}dx = \int \left\{\frac{5}{2}\cdot\frac{2x+2}{x^2+2x+10} - \frac{3}{x^2+2x+10}\right\}dx = \int \left\{\frac{5}{2}\cdot\frac{2x+2}{x^2+2x+10} - \frac{3}{(x+1)^2+9}\right\}dx

=52log⁡(x2+2x+10)−tan⁡−1x+13+C(C は積分定数)= \frac{5}{2}\log(x^2+2x+10) - \tan^{-1}\frac{x+1}{3} + C \qquad (\text{C \text{は積分定数}})

(6)

1x3+1\frac{1}{x^3+1}

∫1x3+1dx=∫1(x+1)(x2−x+1)dx=∫13(1x+1+−x+2x2−x+1)dx\int \frac{1}{x^3+1}dx = \int \frac{1}{(x+1)(x^2-x+1)}dx = \int \frac{1}{3}\left(\frac{1}{x+1} + \frac{-x+2}{x^2-x+1}\right)dx

=∫13(1x+1−2x−1x2−x+1⋅12+32x2−x+1)dx=∫13(1x+1−2x−1x2−x+1⋅12+32⋅1(x−12)2+34)dx= \int \frac{1}{3}\left(\frac{1}{x+1} - \frac{2x-1}{x^2-x+1}\cdot\frac{1}{2} + \frac{\frac{3}{2}}{x^2-x+1}\right)dx = \int \frac{1}{3}\left(\frac{1}{x+1} - \frac{2x-1}{x^2-x+1}\cdot\frac{1}{2} + \frac{3}{2}\cdot\frac{1}{\left(x-\frac{1}{2}\right)^2+\frac{3}{4}}\right)dx

=13{log⁡∣x+1∣−12log⁡∣x2−x+1∣+3tan⁡−1x−1232}+C= \frac{1}{3}\left\{\log|x+1| - \frac{1}{2}\log|x^2-x+1| + \sqrt{3}\tan^{-1}\frac{x-\frac{1}{2}}{\frac{\sqrt{3}}{2}}\right\} + C

=16log⁡(x+1)2x2−x+1+13tan⁡−12x−13+C(C は積分定数)= \frac{1}{6}\log\frac{(x+1)^2}{x^2-x+1} + \frac{1}{\sqrt{3}}\tan^{-1}\frac{2x-1}{\sqrt{3}} + C \qquad (\text{C \text{は積分定数}})

解説: r31bn1z