## 1. 問題の内容

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1. 問題の内容

与えられた積分問題を解く。具体的には、以下の8つの積分を計算する。

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

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

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2. 解き方の手順

各積分について、以下の手順で解いていく。

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

部分分数分解を行う。

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

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

$x = -1$のとき、$1 = -A \implies A = -1$

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

よって、

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

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

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

部分分数分解を行う。

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

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

$x = 1$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、

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

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

部分分数分解を行う。

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

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

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

$B = 2$

$A+B = 3 \implies A = 1$

$A+C = 0 \implies C = -1$

よって、

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

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

部分分数分解を行う。

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

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

$x = -1$のとき、$1 = B(1) \implies B = 1$

$x = -2$のとき、$1 = C(-1)^2 \implies C = 1$

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

よって、

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

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

部分分数分解を行う。

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

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

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

$5A = 5 \implies A = 1$

$4A+C = 1 \implies C = -3$

$A+B = 0 \implies B = -1$

よって、

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

$x^2+4x+5 = (x+2)^2+1$より、$u = x+2$とすると、$x+3 = u+1$

$\int \frac{x+3}{x^2+4x+5}dx = \int \frac{u+1}{u^2+1}du = \int \frac{u}{u^2+1}du + \int \frac{1}{u^2+1}du = \frac{1}{2}\ln(u^2+1) + \arctan(u) + C_1 = \frac{1}{2}\ln(x^2+4x+5) + \arctan(x+2) + C_1$

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

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

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

部分分数分解を行う。

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

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

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

$A+B = 0$

$-A+B+C = 0$

$A+C = 1$

$B = -A$

$-A-A+C = 0 \implies -2A+C = 0 \implies C = 2A$

$A+2A = 1 \implies 3A = 1 \implies A = \frac{1}{3}$

$B = -\frac{1}{3}$

$C = \frac{2}{3}$

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

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

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

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

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

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

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

$\int \frac{1}{(x^2+2)^2} dx$. Let $x = \sqrt{2}\tan\theta$

Then $dx = \sqrt{2}\sec^2\theta d\theta$

$\int \frac{\sqrt{2}\sec^2\theta}{(2\tan^2\theta+2)^2} d\theta = \frac{\sqrt{2}}{4} \int \frac{\sec^2\theta}{\sec^4\theta} d\theta = \frac{\sqrt{2}}{4} \int \cos^2\theta d\theta = \frac{\sqrt{2}}{4} \int \frac{1+\cos2\theta}{2} d\theta = \frac{\sqrt{2}}{8} (\theta + \frac{1}{2}\sin2\theta) = \frac{\sqrt{2}}{8} \left(\arctan\left(\frac{x}{\sqrt{2}}\right) + \frac{x\sqrt{2}}{x^2+2}\right)$

So the result is

$\int \frac{x^4}{(x^2+2)^2}dx = x - \frac{4}{\sqrt{2}} \arctan\left(\frac{x}{\sqrt{2}}\right) + 4 \left(\frac{\sqrt{2}}{8} \left(\arctan\left(\frac{x}{\sqrt{2}}\right) + \frac{x\sqrt{2}}{x^2+2}\right)\right) = x - \sqrt{2} \arctan\left(\frac{x}{\sqrt{2}}\right) + \frac{\sqrt{2}}{2}\left(\arctan\left(\frac{x}{\sqrt{2}}\right) + \frac{x\sqrt{2}}{x^2+2}\right)$

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

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

部分分数分解を行う。

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

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

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

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

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

よって、

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

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

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

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