第4章 4.8

(1)

12x2+2x−12\frac{1}{2x^2+2x-12}

∫12x2+2x−12dx=12∫1x2+x−6dx=12∫1(x−2)(x+3)dx=12∫(1x−2−1x+3)⋅15dx\int \frac{1}{2x^2+2x-12}dx=\frac{1}{2}\int \frac{1}{x^2+x-6}dx=\frac{1}{2}\int \frac{1}{(x-2)(x+3)}dx=\frac{1}{2}\int \left(\frac{1}{x-2}-\frac{1}{x+3}\right)\cdot \frac{1}{5}dx

=110log⁡∣x−2x+3∣+C=\frac{1}{10}\log\left|\frac{x-2}{x+3}\right|+C (Cは積分定数)

(2)

xx4+4x2+8\frac{x}{x^4+4x^2+8}

∫xx4+4x2+8dx=∫x(x2+2)2+4dx=∫2x(x2+2)2+4⋅12dx=14tan⁡−1x2+22+C\int \frac{x}{x^4+4x^2+8}dx=\int \frac{x}{(x^2+2)^2+4}dx=\int \frac{2x}{(x^2+2)^2+4}\cdot \frac{1}{2}dx=\frac{1}{4}\tan^{-1}\frac{x^2+2}{2}+C (Cは積分定数)

(3)

ex1+e2x\frac{e^x}{1+e^{2x}}

∫ex1+e2xdxex=t  とおくとexdx=dtこれから\int \frac{e^x}{1+e^{2x}}dx \qquad e^x=t\ \ \text{とおくと} \qquad e^xdx=dt \qquad \text{これから}

=∫11+t2dt=tan⁡−1t+C元に戻して=\int \frac{1}{1+t^2}dt=\tan^{-1}t+C \qquad \text{元に戻して}

=tan⁡−1ex+C=\tan^{-1}e^x+C (Cは積分定数)

(4)

x4x2−9\frac{x^4}{x^2-9}

∫x4x2−9dx=∫(x2+9)+81x2−9dx=13x3+9x+816log⁡∣x−3x+3∣+C\int \frac{x^4}{x^2-9}dx=\int (x^2+9)+\frac{81}{x^2-9}dx=\frac{1}{3}x^3+9x+\frac{81}{6}\log\left|\frac{x-3}{x+3}\right|+C

=13x3+9x+272log⁡∣x−3x+3∣+C=\frac{1}{3}x^3+9x+\frac{27}{2}\log\left|\frac{x-3}{x+3}\right|+C (Cは積分定数)

(5)

15−4x−x2\frac{1}{\sqrt{5-4x-x^2}}

∫15−4x−x2 dx=∫19−(x+2)2 dx=sin⁡−1x+23+C(C は積分定数)\int \frac{1}{\sqrt{5-4x-x^2}}\,dx = \int \frac{1}{\sqrt{9-(x+2)^2}}\,dx = \sin^{-1}\frac{x+2}{3} + C \qquad \text{(C \text{は積分定数})}

(6)

1x2+6x+8\frac{1}{\sqrt{x^2+6x+8}}

∫1x2+6x+8 dx=∫1(x+3)2−1 dx=log⁡∣x+3+x2+6x+8∣+C(C は積分定数)\int \frac{1}{\sqrt{x^2+6x+8}}\,dx = \int \frac{1}{\sqrt{(x+3)^2-1}}\,dx = \log\left|x+3+\sqrt{x^2+6x+8}\right| + C \qquad \text{(C \text{は積分定数})}

解説: r31bn1z