以下の積分を計算します。 (1) $\int \frac{dx}{x(x+1)^2}$ (3) $\int_0^{\frac{\sqrt{3}}{2}} \frac{dx}{\sqrt{1-x^2}}$ (4) $\int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \frac{dx}{\sin x}$ (5) $\int_0^{1} \frac{dx}{x^3+1}$

わかりました。画像に示された積分問題を解きます。

1. 問題の内容

以下の積分を計算します。

(1) $\int \frac{dx}{x(x+1)^2}$

(3) $\int_0^{\frac{\sqrt{3}}{2}} \frac{dx}{\sqrt{1-x^2}}$

(4) $\int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \frac{dx}{\sin x}$

(5) $\int_0^{1} \frac{dx}{x^3+1}$

2. 解き方の手順

(1) $\int \frac{dx}{x(x+1)^2}$

部分分数分解を行います。

$\frac{1}{x(x+1)^2} = \frac{A}{x} + \frac{B}{x+1} + \frac{C}{(x+1)^2}$

$1 = A(x+1)^2 + Bx(x+1) + Cx$

$x = 0$のとき、$1 = A$

$x = -1$のとき、$1 = -C$ より $C = -1$

$1 = (x+1)^2 + Bx(x+1) - x = x^2+2x+1 + Bx^2+Bx - x = (1+B)x^2 + (1+B)x + 1$

$1+B = 0$ より $B = -1$

よって、

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

(3) $\int_0^{\frac{\sqrt{3}}{2}} \frac{dx}{\sqrt{1-x^2}}$

$\int \frac{dx}{\sqrt{1-x^2}} = \arcsin(x) + C$ なので、

$\int_0^{\frac{\sqrt{3}}{2}} \frac{dx}{\sqrt{1-x^2}} = [\arcsin(x)]_0^{\frac{\sqrt{3}}{2}} = \arcsin(\frac{\sqrt{3}}{2}) - \arcsin(0) = \frac{\pi}{3} - 0 = \frac{\pi}{3}$

(4) $\int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \frac{dx}{\sin x}$

$\int \frac{1}{\sin x} dx = \int \csc x dx = -\ln|\csc x + \cot x| + C$

$\int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \frac{dx}{\sin x} = [-\ln|\csc x + \cot x|]_{\frac{\pi}{3}}^{\frac{\pi}{2}} = -\ln|\csc \frac{\pi}{2} + \cot \frac{\pi}{2}| + \ln|\csc \frac{\pi}{3} + \cot \frac{\pi}{3}|$

$= -\ln|1 + 0| + \ln|\frac{2}{\sqrt{3}} + \frac{1}{\sqrt{3}}| = \ln|\frac{3}{\sqrt{3}}| = \ln|\sqrt{3}| = \frac{1}{2} \ln 3$

(5) $\int_0^{1} \frac{dx}{x^3+1}$

$x^3 + 1 = (x+1)(x^2-x+1)$

$\frac{1}{x^3+1} = \frac{A}{x+1} + \frac{Bx+C}{x^2-x+1}$

$1 = A(x^2-x+1) + (Bx+C)(x+1) = (A+B)x^2 + (-A+B+C)x + (A+C)$

$A+B = 0$, $-A+B+C = 0$, $A+C = 1$

$B = -A$, $C = 1-A$

$-A - A + 1 - A = 0$ より $-3A = -1$, $A = \frac{1}{3}$

$B = -\frac{1}{3}$, $C = \frac{2}{3}$

$\int \frac{dx}{x^3+1} = \int (\frac{\frac{1}{3}}{x+1} + \frac{-\frac{1}{3}x + \frac{2}{3}}{x^2-x+1}) dx = \frac{1}{3} \int \frac{1}{x+1} dx + \frac{1}{3} \int \frac{-x+2}{x^2-x+1} dx$

$= \frac{1}{3} \ln |x+1| + \frac{1}{3} \int \frac{-\frac{1}{2}(2x-1) + \frac{3}{2}}{x^2-x+1} dx = \frac{1}{3} \ln |x+1| - \frac{1}{6} \ln |x^2-x+1| + \frac{1}{2} \int \frac{1}{x^2-x+1} dx$

$x^2-x+1 = (x-\frac{1}{2})^2 + \frac{3}{4}$

$\int \frac{1}{x^2-x+1} dx = \int \frac{1}{(x-\frac{1}{2})^2 + \frac{3}{4}} dx = \frac{2}{\sqrt{3}} \arctan(\frac{2x-1}{\sqrt{3}})$

$\int_0^1 \frac{dx}{x^3+1} = [\frac{1}{3} \ln |x+1| - \frac{1}{6} \ln |x^2-x+1| + \frac{1}{\sqrt{3}} \arctan(\frac{2x-1}{\sqrt{3}})]_0^1 = \frac{1}{3} \ln 2 - \frac{1}{6} \ln 1 + \frac{1}{\sqrt{3}} \arctan(\frac{1}{\sqrt{3}}) - (\frac{1}{3} \ln 1 - \frac{1}{6} \ln 1 + \frac{1}{\sqrt{3}} \arctan(\frac{-1}{\sqrt{3}})) = \frac{1}{3} \ln 2 + \frac{1}{\sqrt{3}} \frac{\pi}{6} - \frac{1}{\sqrt{3}} (-\frac{\pi}{6}) = \frac{1}{3} \ln 2 + \frac{\pi}{3\sqrt{3}}$

3. 最終的な答え

(1) $\ln|\frac{x}{x+1}| + \frac{1}{x+1} + C$

(3) $\frac{\pi}{3}$

(4) $\frac{1}{2} \ln 3$

(5) $\frac{1}{3} \ln 2 + \frac{\pi}{3\sqrt{3}}$