与えられた複数の関数について、不定積分を求める問題です。$a, b > 0$ は定数です。

1. 問題の内容

与えられた複数の関数について、不定積分を求める問題です。$a, b > 0$ は定数です。

2. 解き方の手順

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

部分分数分解を行います。

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

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

$x = 2$ のとき $1 = 4A$ より $A = \frac{1}{4}$

$x = -2$ のとき $1 = -4B$ より $B = -\frac{1}{4}$

よって、

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

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

これは $\int \frac{1}{x^2 + a^2} dx = \frac{1}{a} \arctan \frac{x}{a} + C$ の形を利用します。

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

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

$u = x^2$ と置換すると $du = 2x dx$ より $x dx = \frac{1}{2} du$

$\int \frac{x}{x^4 + 1} dx = \int \frac{1}{u^2 + 1} \cdot \frac{1}{2} du = \frac{1}{2} \int \frac{1}{u^2 + 1} du = \frac{1}{2} \arctan u + C = \frac{1}{2} \arctan (x^2) + C$

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

$x^2 + 2x + 2 = (x+1)^2 + 1$ と変形できます。

$u = x+1$ と置換すると $du = dx$

$\int \frac{1}{x^2 + 2x + 2} dx = \int \frac{1}{(x+1)^2 + 1} dx = \int \frac{1}{u^2 + 1} du = \arctan u + C = \arctan (x+1) + C$

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

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

$u = x + \frac{1}{2}$ と置換すると $du = dx$

$\int \frac{3}{x^2 + x + 1} dx = 3 \int \frac{1}{(x + \frac{1}{2})^2 + (\frac{\sqrt{3}}{2})^2} dx = 3 \int \frac{1}{u^2 + (\frac{\sqrt{3}}{2})^2} du = 3 \cdot \frac{2}{\sqrt{3}} \arctan \frac{u}{\frac{\sqrt{3}}{2}} + C = 2 \sqrt{3} \arctan \frac{2x+1}{\sqrt{3}} + C$

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

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

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

(7) $\int \frac{1}{\sqrt{9 - x^2}} dx$

これは $\int \frac{1}{\sqrt{a^2 - x^2}} dx = \arcsin \frac{x}{a} + C$ の形を利用します。

$\int \frac{1}{\sqrt{9 - x^2}} dx = \int \frac{1}{\sqrt{3^2 - x^2}} dx = \arcsin \frac{x}{3} + C$

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

これは $\int \frac{1}{\sqrt{x^2 + a^2}} dx = \operatorname{arcsinh} \frac{x}{a} + C = \ln(x + \sqrt{x^2 + a^2}) + C$ の形を利用します。

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

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

$u = 1+x$ と置換すると $x = u-1$ であり $du = dx$

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

(10) $\int \sqrt{5x + 4} dx$

$u = 5x + 4$ と置換すると $du = 5 dx$ より $dx = \frac{1}{5} du$

$\int \sqrt{5x + 4} dx = \int \sqrt{u} \cdot \frac{1}{5} du = \frac{1}{5} \int u^{\frac{1}{2}} du = \frac{1}{5} \cdot \frac{2}{3} u^{\frac{3}{2}} + C = \frac{2}{15} (5x+4)^{\frac{3}{2}} + C$

(11) $\int x \sqrt{x-5} dx$

$u = x-5$ と置換すると $x = u+5$ であり $du = dx$

$\int x \sqrt{x-5} dx = \int (u+5) \sqrt{u} du = \int (u^{\frac{3}{2}} + 5 u^{\frac{1}{2}}) du = \frac{2}{5} u^{\frac{5}{2}} + 5 \cdot \frac{2}{3} u^{\frac{3}{2}} + C = \frac{2}{5} (x-5)^{\frac{5}{2}} + \frac{10}{3} (x-5)^{\frac{3}{2}} + C = \frac{2}{5}(x-5)^2 \sqrt{x-5} + \frac{10}{3}(x-5)\sqrt{x-5} + C = \frac{2}{15}(3x^2 - 15)\sqrt{x-5} + C = \frac{2}{15}(3x+25)(x-5)^{\frac{3}{2}} + C$

(12) $\int x \sqrt{a^2 - x^2} dx$

$u = a^2 - x^2$ と置換すると $du = -2x dx$ より $x dx = -\frac{1}{2} du$

$\int x \sqrt{a^2 - x^2} dx = \int \sqrt{u} \cdot (-\frac{1}{2}) du = -\frac{1}{2} \int u^{\frac{1}{2}} du = -\frac{1}{2} \cdot \frac{2}{3} u^{\frac{3}{2}} + C = -\frac{1}{3} (a^2 - x^2)^{\frac{3}{2}} + C$

(13) $\int \frac{x}{\sqrt{a^2 - x^2}} dx$

$u = a^2 - x^2$ と置換すると $du = -2x dx$ より $x dx = -\frac{1}{2} du$

$\int \frac{x}{\sqrt{a^2 - x^2}} dx = \int \frac{1}{\sqrt{u}} \cdot (-\frac{1}{2}) du = -\frac{1}{2} \int u^{-\frac{1}{2}} du = -\frac{1}{2} \cdot 2 u^{\frac{1}{2}} + C = - \sqrt{a^2 - x^2} + C$

(14) $\int \frac{1}{\sqrt{x(1-x)}} dx$

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

$u = x - \frac{1}{2}$ と置換すると $du = dx$

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

(15) $\int \sqrt{x^2 + a^2} dx$

これは $\int \sqrt{x^2 + a^2} dx = \frac{1}{2} x \sqrt{x^2 + a^2} + \frac{1}{2} a^2 \operatorname{arcsinh} \frac{x}{a} + C = \frac{1}{2}x\sqrt{x^2+a^2} + \frac{1}{2}a^2 \ln(x+\sqrt{x^2+a^2}) + C$

(16) $\int x \log(x + \sqrt{1 + x^2}) dx$

部分積分を行います。$u = \log(x + \sqrt{1+x^2})$, $dv = x dx$

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

$v = \frac{1}{2}x^2$

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

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

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

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

$\frac{x^2 + 5x + 4}{(x+2)(x^2 + 2x + 2)} = \frac{(x+1)(x+4)}{(x+2)(x^2 + 2x + 2)} = \frac{A}{x+2} + \frac{Bx + C}{x^2+2x+2}$

$(x+1)(x+4) = A(x^2+2x+2) + (Bx+C)(x+2)$

$x = -2$ のとき $(-1)(2) = A(4-4+2)$ より $2A = -2$ なので $A = -1$

$x^2+5x+4 = -x^2-2x-2 + Bx^2 + 2Bx + Cx + 2C$

$2x^2+7x+6 = Bx^2 + (2B+C)x + 2C$

$B = 2$, $2B+C = 7$ より $4+C = 7$ なので $C = 3$

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

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

(18) $\int \frac{1}{x \sqrt{x-1}} dx$

$u = \sqrt{x-1}$ と置換すると $u^2 = x-1$ より $x = u^2 + 1$ であり $2u du = dx$

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

(19) $\int \frac{e^x}{e^{2x} - 3e^x + 2} dx$

$u = e^x$ と置換すると $du = e^x dx$

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

(20) $\int \frac{\cos x}{\sin^2 x + 3 \sin x} dx$

$u = \sin x$ と置換すると $du = \cos x dx$

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

(21) $\int \frac{1}{a^2 \cos^2 x + b^2 \sin^2 x} dx$

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

$u = \tan x$ と置換すると $du = \sec^2 x dx$

$\int \frac{\sec^2 x}{a^2 + b^2 \tan^2 x} dx = \int \frac{1}{a^2 + b^2 u^2} du = \frac{1}{b^2} \int \frac{1}{(\frac{a}{b})^2 + u^2} du = \frac{1}{b^2} \cdot \frac{b}{a} \arctan(\frac{bu}{a}) + C = \frac{1}{ab} \arctan(\frac{b}{a} \tan x) + C$

3. 最終的な答え

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

(2) $\frac{1}{\sqrt{3}} \arctan \frac{x}{\sqrt{3}} + C$

(3) $\frac{1}{2} \arctan (x^2) + C$

(4) $\arctan (x+1) + C$

(5) $2 \sqrt{3} \arctan \frac{2x+1}{\sqrt{3}} + C$

(6) $x - 2 \arctan \frac{x}{2} + C$

(7) $\arcsin \frac{x}{3} + C$

(8) $\ln(x + \sqrt{x^2 + 2}) + C$

(9) $\frac{2}{3} (x-2)\sqrt{1+x} + C$

(10) $\frac{2}{15} (5x+4)^{\frac{3}{2}} + C$

(11) $\frac{2}{15}(3x+25)(x-5)^{\frac{3}{2}} + C$

(12) $-\frac{1}{3} (a^2 - x^2)^{\frac{3}{2}} + C$

(13) $- \sqrt{a^2 - x^2} + C$

(14) $\arcsin (2x - 1) + C$

(15) $\frac{1}{2}x\sqrt{x^2+a^2} + \frac{1}{2}a^2 \ln(x+\sqrt{x^2+a^2}) + C$

(16) $\frac{1}{2} x^2 \log(x + \sqrt{1 + x^2}) - \frac{1}{4} x \sqrt{1+x^2} + \frac{1}{4} \ln (x + \sqrt{1+x^2}) + C$

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

(18) $2 \arctan \sqrt{x-1} + C$

(19) $\ln|\frac{e^x-2}{e^x-1}| + C$

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

(21) $\frac{1}{ab} \arctan(\frac{b}{a} \tan x) + C$