第4章 4.5

(1)

x2x2+1\dfrac{x^2}{x^2+1}

∫x2x2+1 dx=∫1−1x2+1 dx=x−tan⁡−1x+C\int \frac{x^2}{x^2+1}\,dx = \int 1-\frac{1}{x^2+1}\,dx = x-\tan^{-1}x+C (C は積分定数)

(2)

2x−1x2+1\dfrac{2x-1}{x^2+1}

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

(3)

2x+5x2+4x+5\dfrac{2x+5}{x^2+4x+5}

∫2x+5x2+4x+5 dx=∫{2x+4x2+4x+5+1x2+4x+5}dx=∫{2x+4x2+4x+5+1(x+2)2+1}dx\int \frac{2x+5}{x^2+4x+5}\,dx = \int \left\{\frac{2x+4}{x^2+4x+5}+\frac{1}{x^2+4x+5}\right\}dx = \int \left\{\frac{2x+4}{x^2+4x+5}+\frac{1}{(x+2)^2+1}\right\}dx

=log⁡(x2+4x+5)+tan⁡−1(x+1)+C=\log(x^2+4x+5)+\tan^{-1}(x+1)+C (C は積分定数)

(4)

x2x6+4\dfrac{x^2}{x^6+4}

∫x2x6+4 dxx3=tとおくと3x2 dx=dtx2 dx=13dtこれから\int \frac{x^2}{x^6+4}\,dx \qquad x^3=t \quad \text{とおくと} \quad 3x^2\,dx=dt \quad x^2\,dx=\frac{1}{3}dt \quad \text{これから}

=∫13⋅1t2+4 dt=16tan⁡−1t2+C元に戻して=\int \frac{1}{3}\cdot\frac{1}{t^2+4}\,dt = \frac{1}{6}\tan^{-1}\frac{t}{2}+C \qquad \text{元に戻して}

=16tan⁡−1x32+C=\frac{1}{6}\tan^{-1}\frac{x^3}{2}+C (C は積分定数)

解説: r31bn1z