与えられた6つの関数を積分する問題です。 (1) $\frac{x^2}{(2x-1)^2}$ (2) $xe^{-x^2}$ (3) $\frac{1}{x^2+4x+9}$ (4) $\frac{1}{\sqrt{x^2+4x+7}}$ (5) $\frac{1}{\sqrt{3+2x-x^2}}$ (6) $\frac{e^x}{1+e^{2x}}$

1. 問題の内容

与えられた6つの関数を積分する問題です。

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

(2) $xe^{-x^2}$

(3) $\frac{1}{x^2+4x+9}$

(4) $\frac{1}{\sqrt{x^2+4x+7}}$

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

(6) $\frac{e^x}{1+e^{2x}}$

2. 解き方の手順

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

まず、$\frac{x^2}{(2x-1)^2}$ を部分分数分解します。

$\frac{x^2}{(2x-1)^2} = \frac{A}{2x-1} + \frac{B}{(2x-1)^2} + C$ と置きます。

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

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

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

係数を比較すると、

$4C = 1 \Rightarrow C = \frac{1}{4}$

$2A - 4C = 0 \Rightarrow 2A = 4C = 1 \Rightarrow A = \frac{1}{2}$

$-A + B + C = 0 \Rightarrow B = A - C = \frac{1}{2} - \frac{1}{4} = \frac{1}{4}$

したがって、$\frac{x^2}{(2x-1)^2} = \frac{1/2}{2x-1} + \frac{1/4}{(2x-1)^2} + \frac{1}{4}$ です。

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

$= \frac{1}{4} \ln |2x-1| - \frac{1}{8(2x-1)} + \frac{1}{4} x + C$

(2) $xe^{-x^2}$

$u = -x^2$ と置くと、$\frac{du}{dx} = -2x$ より $x dx = -\frac{1}{2} du$

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

(3) $\frac{1}{x^2+4x+9}$

$x^2 + 4x + 9 = (x+2)^2 + 5$ なので、

$\int \frac{1}{x^2+4x+9} dx = \int \frac{1}{(x+2)^2 + 5} dx$

$u = x+2$ と置くと、$du = dx$

$\int \frac{1}{u^2 + 5} du = \frac{1}{\sqrt{5}} \arctan \frac{u}{\sqrt{5}} + C = \frac{1}{\sqrt{5}} \arctan \frac{x+2}{\sqrt{5}} + C$

(4) $\frac{1}{\sqrt{x^2+4x+7}}$

$x^2+4x+7 = (x+2)^2 + 3$

$\int \frac{1}{\sqrt{x^2+4x+7}} dx = \int \frac{1}{\sqrt{(x+2)^2 + 3}} dx$

$u = x+2$ と置くと、$du = dx$

$\int \frac{1}{\sqrt{u^2 + 3}} du = \operatorname{arcsinh}\left(\frac{u}{\sqrt{3}}\right) + C = \ln(u + \sqrt{u^2 + 3}) + C = \ln(x+2 + \sqrt{x^2+4x+7}) + C$

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

$3+2x-x^2 = 4 - (x-1)^2$

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

$u = x-1$ と置くと、$du = dx$

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

(6) $\frac{e^x}{1+e^{2x}}$

$u = e^x$ と置くと、$du = e^x dx$

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

3. 最終的な答え

(1) $\int \frac{x^2}{(2x-1)^2} dx = \frac{1}{4} \ln |2x-1| - \frac{1}{8(2x-1)} + \frac{1}{4} x + C$

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

(3) $\int \frac{1}{x^2+4x+9} dx = \frac{1}{\sqrt{5}} \arctan \frac{x+2}{\sqrt{5}} + C$

(4) $\int \frac{1}{\sqrt{x^2+4x+7}} dx = \ln(x+2 + \sqrt{x^2+4x+7}) + C$

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

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