第8章 8.5

(1)

∫4(x−1)(x−2)(x+2)dx=∫(1x−2+3x+2)dx=log⁡∣x−2∣+log⁡∣(x+2)3∣+C=log⁡∣(x−2)(x+2)3∣+C\int \frac{4(x-1)}{(x-2)(x+2)}dx = \int \left(\frac{1}{x-2}+\frac{3}{x+2}\right)dx = \log|x-2|+\log|(x+2)^3|+C = \log|(x-2)(x+2)^3|+C

(2)

∫(x2+1)2x3dx=∫x4+2x2+1x3dx=∫(x+2x+1x3)dx=12x2+2log⁡x−12x2+C\int \frac{(x^2+1)^2}{x^3}dx = \int \frac{x^4+2x^2+1}{x^3}dx = \int \left(x+\frac{2}{x}+\frac{1}{x^3}\right)dx = \frac{1}{2}x^2+2\log x - \frac{1}{2x^2}+C

(3)

∫t3t2−3t+2dt=∫(t+3+7t−6t2−3t+2)dt=∫(t+3+7t−6(t−1)(t−2))dt=∫(t+3+−1t−1+8t−2)dx\int \frac{t^3}{t^2-3t+2}dt = \int \left(t+3+\frac{7t-6}{t^2-3t+2}\right)dt = \int \left(t+3+\frac{7t-6}{(t-1)(t-2)}\right)dt = \int \left(t+3+\frac{-1}{t-1}+\frac{8}{t-2}\right)dx

=12t2+3t−log⁡(t−1)+8log⁡(t−2)+C=12t2+3t+log⁡(t−2)8∣t−1∣+C= \frac{1}{2}t^2+3t-\log(t-1)+8\log(t-2)+C = \frac{1}{2}t^2+3t+\log\frac{(t-2)^8}{|t-1|}+C

(4)

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

=∫(1+3(x+1)2+−2x+1)dx=x−3x+1−2log⁡∣x+1∣+C= \int \left(1+\frac{3}{(x+1)^2}+\frac{-2}{x+1}\right)dx = x-\frac{3}{x+1}-2\log|x+1|+C

解説: r31bn1z