高専の数学解説

第4章 4.13

(1)

∫x(x2+a2)2 dx=−12(x2+a2)\int \frac{x}{(x^2+a^2)^2}\,dx = -\frac{1}{2(x^2+a^2)}

∫x(x2+a2)2 dx=∫2x(x2+a2)2⋅12 dx=−12(x2+a2)+C(C は積分定数)\int \frac{x}{(x^2+a^2)^2}\,dx = \int \frac{2x}{(x^2+a^2)^2}\cdot\frac{1}{2}\,dx = -\frac{1}{2(x^2+a^2)}+C \qquad \text{(C \text{は積分定数})}

(2)

∫x2(x2+a2)2 dx=12(−xx2+a2+1atan⁡−1xa)\int \frac{x^2}{(x^2+a^2)^2}\,dx = \frac{1}{2}\left(-\frac{x}{x^2+a^2}+\frac{1}{a}\tan^{-1}\frac{x}{a}\right)

∫x2(x2+a2)2 dx=∫xx(x2+a2)2 dx=x(−12(x2+a2))+12∫1x2+a2 dx\int \frac{x^2}{(x^2+a^2)^2}\,dx = \int x\frac{x}{(x^2+a^2)^2}\,dx = x\left(-\frac{1}{2(x^2+a^2)}\right)+\frac{1}{2}\int \frac{1}{x^2+a^2}\,dx

=−x2(x2+a2)+12⋅1atan⁡−1xa+C=12(−xx2+a2+1atan⁡−1xa)+C(C は積分定数)= -\frac{x}{2(x^2+a^2)}+\frac{1}{2}\cdot\frac{1}{a}\tan^{-1}\frac{x}{a}+C = \frac{1}{2}\left(-\frac{x}{x^2+a^2}+\frac{1}{a}\tan^{-1}\frac{x}{a}\right)+C \qquad \text{(C \text{は積分定数})}

(3)

∫1(x2+a2)2 dx=12a2(xx2+a2+1atan⁡−1xa)\int \frac{1}{(x^2+a^2)^2}\,dx = \frac{1}{2a^2}\left(\frac{x}{x^2+a^2}+\frac{1}{a}\tan^{-1}\frac{x}{a}\right)

∫1(x2+a2)2 dx=∫1a2{1x2+a2−x2(x2+a2)2}dx\int \frac{1}{(x^2+a^2)^2}\,dx = \int \frac{1}{a^2}\left\{\frac{1}{x^2+a^2}-\frac{x^2}{(x^2+a^2)^2}\right\}dx

=1a2{1atan⁡−1xa−12(−xx2+a2+1atan⁡−1xa)}+C=12a2(xx2+a2+1atan⁡−1xa)+C= \frac{1}{a^2}\left\{\frac{1}{a}\tan^{-1}\frac{x}{a}-\frac{1}{2}\left(-\frac{x}{x^2+a^2}+\frac{1}{a}\tan^{-1}\frac{x}{a}\right)\right\}+C = \frac{1}{2a^2}\left(\frac{x}{x^2+a^2}+\frac{1}{a}\tan^{-1}\frac{x}{a}\right)+C

（C は積分定数）

(4)

In=12(n−1)a2{x(x2+a2)n−1+(2n−3)In−1}(n≥2)I_n = \frac{1}{2(n-1)a^2}\left\{\frac{x}{(x^2+a^2)^{n-1}}+(2n-3)I_{n-1}\right\} \qquad (n\ge 2)

In=∫1(x2+a2)n dx=∫{1a2(x2+a2)n−1−x2a2(x2+a2)n}dx=∫1a2⋅1(x2+a2)n−1 dx−1a2∫x2(x2+a2)n dxI_n = \int \frac{1}{(x^2+a^2)^n}\,dx = \int \left\{\frac{1}{a^2(x^2+a^2)^{n-1}}-\frac{x^2}{a^2(x^2+a^2)^n}\right\}dx = \int \frac{1}{a^2}\cdot\frac{1}{(x^2+a^2)^{n-1}}\,dx - \frac{1}{a^2}\int \frac{x^2}{(x^2+a^2)^n}\,dx

ここで z=∫x2(x2+a2)n dxz = \int \frac{x^2}{(x^2+a^2)^n}\,dx とすると

z=∫x(x2+a2)n⋅x dx=−12(n−1)⋅1(x2+a2)n−1⋅x+∫12(n−1)⋅1(x2+a2)n−1 dxz = \int \frac{x}{(x^2+a^2)^n}\cdot x\,dx = -\frac{1}{2(n-1)}\cdot\frac{1}{(x^2+a^2)^{n-1}}\cdot x + \int \frac{1}{2(n-1)}\cdot\frac{1}{(x^2+a^2)^{n-1}}\,dx

=−12(n−1){x(x2+a2)n−1−∫1(x2+a2)n−1 dx}=−12(n−1){x(x2+a2)n−1−In−1}= -\frac{1}{2(n-1)}\left\{\frac{x}{(x^2+a^2)^{n-1}} - \int \frac{1}{(x^2+a^2)^{n-1}}\,dx\right\} = -\frac{1}{2(n-1)}\left\{\frac{x}{(x^2+a^2)^{n-1}} - I_{n-1}\right\}

これから

In=∫1a2⋅1(x2+a2)n−1 dx−1a2⋅{−12(n−1)(x(x2+a2)n−1−In−1)}I_n = \int \frac{1}{a^2}\cdot\frac{1}{(x^2+a^2)^{n-1}}\,dx - \frac{1}{a^2}\cdot\left\{-\frac{1}{2(n-1)}\left(\frac{x}{(x^2+a^2)^{n-1}} - I_{n-1}\right)\right\}

=1a2In−1+12(n−1)a2{x(x2+a2)n−1−In−1}=12(n−1)a2{2(n−1)In−1+x(x2+a2)n−1−In−1}= \frac{1}{a^2}I_{n-1} + \frac{1}{2(n-1)a^2}\left\{\frac{x}{(x^2+a^2)^{n-1}} - I_{n-1}\right\} = \frac{1}{2(n-1)a^2}\left\{2(n-1)I_{n-1} + \frac{x}{(x^2+a^2)^{n-1}} - I_{n-1}\right\}

=12(n−1)a2{x(x2+a2)n−1+(2n−3)In−1}= \frac{1}{2(n-1)a^2}\left\{\frac{x}{(x^2+a^2)^{n-1}} + (2n-3)I_{n-1}\right\}

解説: r31bn1z