次の関数の不定積分を求めます。 (13) $f(x) = \frac{1}{x(\log x)^2}$ (14) $f(x) = \frac{1}{3 + \cos x}$ (15) $f(x) = \frac{x}{x^4 - 1}$ (16) $f(x) = \frac{\log x}{x^2}$ (17) $f(x) = \frac{1}{1 + \sin x}$ (18) $f(x) = \tan^{-1} x$

1. 問題の内容

次の関数の不定積分を求めます。

(13) $f(x) = \frac{1}{x(\log x)^2}$

(14) $f(x) = \frac{1}{3 + \cos x}$

(15) $f(x) = \frac{x}{x^4 - 1}$

(16) $f(x) = \frac{\log x}{x^2}$

(17) $f(x) = \frac{1}{1 + \sin x}$

(18) $f(x) = \tan^{-1} x$

2. 解き方の手順

(13) $f(x) = \frac{1}{x(\log x)^2}$

$\log x = t$ と置換すると、$\frac{1}{x} dx = dt$ となる。

$\int \frac{1}{x(\log x)^2} dx = \int \frac{1}{t^2} dt = -\frac{1}{t} + C = -\frac{1}{\log x} + C$

(14) $f(x) = \frac{1}{3 + \cos x}$

$t = \tan \frac{x}{2}$ と置換すると、$\cos x = \frac{1-t^2}{1+t^2}$、 $dx = \frac{2}{1+t^2} dt$ となる。

$\int \frac{1}{3 + \cos x} dx = \int \frac{1}{3 + \frac{1-t^2}{1+t^2}} \cdot \frac{2}{1+t^2} dt = \int \frac{2}{3(1+t^2) + (1-t^2)} dt = \int \frac{2}{4 + 2t^2} dt = \int \frac{1}{2+t^2} dt$

$= \frac{1}{\sqrt{2}} \arctan \frac{t}{\sqrt{2}} + C = \frac{1}{\sqrt{2}} \arctan \frac{\tan \frac{x}{2}}{\sqrt{2}} + C$

(15) $f(x) = \frac{x}{x^4 - 1}$

$\frac{x}{x^4 - 1} = \frac{x}{(x^2 - 1)(x^2 + 1)} = \frac{x}{(x-1)(x+1)(x^2+1)}$

部分分数分解をすると、$\frac{x}{x^4 - 1} = \frac{A}{x-1} + \frac{B}{x+1} + \frac{Cx+D}{x^2+1}$

$x = A(x+1)(x^2+1) + B(x-1)(x^2+1) + (Cx+D)(x^2-1)$

$x = A(x^3 + x^2 + x + 1) + B(x^3 - x^2 + x - 1) + Cx^3 - Cx + Dx^2 - D$

$x = (A+B+C)x^3 + (A-B+D)x^2 + (A+B-C)x + (A-B-D)$

$A+B+C = 0$

$A-B+D = 0$

$A+B-C = 1$

$A-B-D = 0$

$2(A+B) = 1 \implies A+B = \frac{1}{2}$

$2(A-B) = 0 \implies A=B = \frac{1}{4}$

$C = A+B-1 = \frac{1}{2} - 1 = -\frac{1}{2}$

$D = B-A = 0$

$\int \frac{x}{x^4 - 1} dx = \int \frac{1/4}{x-1} + \frac{1/4}{x+1} + \frac{-x/2}{x^2+1} dx = \frac{1}{4} \log |x-1| + \frac{1}{4} \log |x+1| - \frac{1}{4} \log (x^2+1) + C = \frac{1}{4} \log \left| \frac{x^2-1}{x^2+1} \right| + C$

(16) $f(x) = \frac{\log x}{x^2}$

部分積分をする。

$u = \log x, dv = \frac{1}{x^2} dx$

$du = \frac{1}{x} dx, v = -\frac{1}{x}$

$\int \frac{\log x}{x^2} dx = -\frac{\log x}{x} - \int -\frac{1}{x} \cdot \frac{1}{x} dx = -\frac{\log x}{x} + \int \frac{1}{x^2} dx = -\frac{\log x}{x} - \frac{1}{x} + C = -\frac{\log x + 1}{x} + C$

(17) $f(x) = \frac{1}{1 + \sin x}$

$\int \frac{1}{1+\sin x} dx = \int \frac{1-\sin x}{1 - \sin^2 x} dx = \int \frac{1-\sin x}{\cos^2 x} dx = \int (\sec^2 x - \sec x \tan x) dx = \tan x - \sec x + C$

(18) $f(x) = \tan^{-1} x$

部分積分をする。

$u = \tan^{-1} x, dv = dx$

$du = \frac{1}{1+x^2} dx, v = x$

$\int \tan^{-1} x dx = x \tan^{-1} x - \int \frac{x}{1+x^2} dx = x \tan^{-1} x - \frac{1}{2} \log(1+x^2) + C$

3. 最終的な答え

(13) $-\frac{1}{\log x} + C$

(14) $\frac{1}{\sqrt{2}} \arctan \frac{\tan \frac{x}{2}}{\sqrt{2}} + C$

(15) $\frac{1}{4} \log \left| \frac{x^2-1}{x^2+1} \right| + C$

(16) $-\frac{\log x + 1}{x} + C$

(17) $\tan x - \sec x + C$

(18) $x \tan^{-1} x - \frac{1}{2} \log(1+x^2) + C$