第4章 4.14

(1)

1(x2+4)2\frac{1}{(x^2+4)^2}

∫1(x2+4)2 dx=18(xx2+4+12tan⁡−1x2)+C\int \frac{1}{(x^2+4)^2}\,dx = \frac{1}{8}\left(\frac{x}{x^2+4} + \frac{1}{2}\tan^{-1}\frac{x}{2}\right) + C (C は積分定数)

(2)

x2(x2+4)2\frac{x^2}{(x^2+4)^2}

∫x2(x2+4)2 dx=12(−xx2+4+12tan⁡−1x2)+C\int \frac{x^2}{(x^2+4)^2}\,dx = \frac{1}{2}\left(-\frac{x}{x^2+4} + \frac{1}{2}\tan^{-1}\frac{x}{2}\right) + C (C は積分定数)

(3)

1(x2+4)3\frac{1}{(x^2+4)^3}

∫1(x2+4)3 dx=116{x(x2+4)2+3⋅18(xx2+4+12tan⁡−1x2)}+C\int \frac{1}{(x^2+4)^3}\,dx = \frac{1}{16}\left\{\frac{x}{(x^2+4)^2} + 3\cdot\frac{1}{8}\left(\frac{x}{x^2+4} + \frac{1}{2}\tan^{-1}\frac{x}{2}\right)\right\} + C

=116{x(x2+4)2+38(xx2+4+12tan⁡−1x2)}+C= \frac{1}{16}\left\{\frac{x}{(x^2+4)^2} + \frac{3}{8}\left(\frac{x}{x^2+4} + \frac{1}{2}\tan^{-1}\frac{x}{2}\right)\right\} + C (C は積分定数)

解説: r31bn1z