与えられた積分問題を解きます。具体的には、以下の8つの不定積分を計算します。 (1) $\int \frac{1}{(x+1)(2x+1)}dx$ (2) $\int \frac{x^2}{x^2-1}dx$ (3) $\int \frac{3x+2}{x^2(x+1)}dx$ (4) $\int \frac{1}{(x+1)(x+2)}dx$ (5) $\int \frac{x+5}{x(x^2+4x+5)}dx$ (6) $\int \frac{1}{x^3+1}dx$ (7) $\int \frac{x^4}{(x^2+2)^2}dx$ (8) $\int \frac{1}{(x-1)(x-2)(x-3)}dx$

1. 問題の内容

与えられた積分問題を解きます。具体的には、以下の8つの不定積分を計算します。

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

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

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

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

(5) $\int \frac{x+5}{x(x^2+4x+5)}dx$

(6) $\int \frac{1}{x^3+1}dx$

(7) $\int \frac{x^4}{(x^2+2)^2}dx$

(8) $\int \frac{1}{(x-1)(x-2)(x-3)}dx$

2. 解き方の手順

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

部分分数分解を行います。

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

$1 = A(2x+1) + B(x+1)$

$x = -1$ のとき、$1 = A(-2+1) + B(0) \implies A = -1$

$x = -\frac{1}{2}$ のとき、$1 = A(0) + B(-\frac{1}{2}+1) \implies 1 = \frac{1}{2}B \implies B = 2$

よって、$\frac{1}{(x+1)(2x+1)} = \frac{-1}{x+1} + \frac{2}{2x+1}$

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

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

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

$\frac{1}{(x-1)(x+1)} = \frac{A}{x-1} + \frac{B}{x+1}$

$1 = A(x+1) + B(x-1)$

$x = 1$ のとき、$1 = 2A \implies A = \frac{1}{2}$

$x = -1$ のとき、$1 = -2B \implies B = -\frac{1}{2}$

よって、$\frac{1}{x^2-1} = \frac{1}{2}\left( \frac{1}{x-1} - \frac{1}{x+1} \right)$

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

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

$\frac{3x+2}{x^2(x+1)} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x+1}$

$3x+2 = Ax(x+1) + B(x+1) + Cx^2$

$3x+2 = Ax^2+Ax + Bx + B + Cx^2$

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

$A+C=0, A+B=3, B=2 \implies A=1, C=-1$

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

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

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

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

$1 = A(x+2) + B(x+1)$

$x=-1$ のとき、$1 = A \implies A=1$

$x=-2$ のとき、$1 = -B \implies B=-1$

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

(5) $\int \frac{x+5}{x(x^2+4x+5)}dx$

$\frac{x+5}{x(x^2+4x+5)} = \frac{A}{x} + \frac{Bx+C}{x^2+4x+5}$

$x+5 = A(x^2+4x+5) + (Bx+C)x$

$x+5 = Ax^2 + 4Ax + 5A + Bx^2 + Cx$

$x+5 = (A+B)x^2 + (4A+C)x + 5A$

$A+B=0, 4A+C=1, 5A=5 \implies A=1, B=-1, C=-3$

$\frac{x+5}{x(x^2+4x+5)} = \frac{1}{x} + \frac{-x-3}{x^2+4x+5}$

$\int \frac{x+5}{x(x^2+4x+5)}dx = \int \left( \frac{1}{x} - \frac{x+3}{x^2+4x+5} \right) dx = \int \frac{1}{x}dx - \frac{1}{2}\int \frac{2x+6}{x^2+4x+5}dx$

$= \ln|x| - \frac{1}{2}\int \frac{2x+4+2}{x^2+4x+5}dx = \ln|x| - \frac{1}{2}\ln|x^2+4x+5| - \int \frac{1}{(x+2)^2+1}dx = \ln|x| - \frac{1}{2}\ln|x^2+4x+5| - \arctan(x+2) + C$

(6) $\int \frac{1}{x^3+1}dx$

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

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

$1 = Ax^2 - Ax + A + Bx^2 + Bx + Cx + C$

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

$A+B=0, -A+B+C=0, A+C=1 \implies A=\frac{1}{3}, B=-\frac{1}{3}, C=\frac{2}{3}$

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

$\int \frac{1}{x^3+1}dx = \frac{1}{3}\int \frac{1}{x+1}dx + \frac{1}{3}\int \frac{-x+2}{x^2-x+1}dx = \frac{1}{3}\ln|x+1| - \frac{1}{6}\int \frac{2x-4}{x^2-x+1}dx = \frac{1}{3}\ln|x+1| - \frac{1}{6}\int \frac{2x-1-3}{x^2-x+1}dx$

$= \frac{1}{3}\ln|x+1| - \frac{1}{6}\ln|x^2-x+1| + \frac{1}{2}\int \frac{1}{x^2-x+1}dx = \frac{1}{3}\ln|x+1| - \frac{1}{6}\ln|x^2-x+1| + \frac{1}{2}\int \frac{1}{(x-\frac{1}{2})^2 + \frac{3}{4}}dx$

$= \frac{1}{3}\ln|x+1| - \frac{1}{6}\ln|x^2-x+1| + \frac{1}{2}\frac{2}{\sqrt{3}}\arctan\left( \frac{x-\frac{1}{2}}{\frac{\sqrt{3}}{2}} \right) + C = \frac{1}{3}\ln|x+1| - \frac{1}{6}\ln|x^2-x+1| + \frac{1}{\sqrt{3}}\arctan\left( \frac{2x-1}{\sqrt{3}} \right) + C$

(7) $\int \frac{x^4}{(x^2+2)^2}dx$

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

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

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

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

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

(8) $\int \frac{1}{(x-1)(x-2)(x-3)}dx$

$\frac{1}{(x-1)(x-2)(x-3)} = \frac{A}{x-1} + \frac{B}{x-2} + \frac{C}{x-3}$

$1 = A(x-2)(x-3) + B(x-1)(x-3) + C(x-1)(x-2)$

$x=1 \implies 1= A(-1)(-2)=2A \implies A=\frac{1}{2}$

$x=2 \implies 1 = B(1)(-1) = -B \implies B = -1$

$x=3 \implies 1 = C(2)(1) = 2C \implies C = \frac{1}{2}$

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

$=\frac{1}{2}(\ln|x-1|+\ln|x-3|) - \ln|x-2| + C = \frac{1}{2} \ln|(x-1)(x-3)| - \ln|x-2| + C = \ln \frac{\sqrt{|(x-1)(x-3)|}}{|x-2|} + C$

3. 最終的な答え

(1) $\ln\left| \frac{2x+1}{x+1} \right| + C$

(2) $x + \frac{1}{2}\ln\left| \frac{x-1}{x+1} \right| + C$

(3) $\ln\left| \frac{x}{x+1} \right| - \frac{2}{x} + C$

(4) $\ln\left| \frac{x+1}{x+2} \right| + C$

(5) $\ln|x| - \frac{1}{2}\ln|x^2+4x+5| - \arctan(x+2) + C$

(6) $\frac{1}{3}\ln|x+1| - \frac{1}{6}\ln|x^2-x+1| + \frac{1}{\sqrt{3}}\arctan\left( \frac{2x-1}{\sqrt{3}} \right) + C$

(7) $x - \sqrt{2}\arctan\left( \frac{x}{\sqrt{2}} \right) + \frac{x}{x^2+2} + C$

(8) $\ln \frac{\sqrt{|(x-1)(x-3)|}}{|x-2|} + C$