## 解答

## 解答

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1. 問題の内容

与えられた7つの積分（不定積分および定積分）を計算します。

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

(2) $\int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \frac{1-\cos^3x}{\cos^2x} dx$

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

(4) $\int_{0}^{1} (e^x + e^{-x})^2 dx$

(5) $\int_{0}^{1} xe^{x^2+1} dx$

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

(7) $\int_{1}^{e} \sqrt{x} \log x dx$

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2. 解き方の手順

**(1) $\int (\sqrt[3]{x^2} - 2\frac{1}{\sqrt[3]{x}} + \frac{1}{x}) dx$**

$\sqrt[3]{x^2} = x^{\frac{2}{3}}$、$\frac{1}{\sqrt[3]{x}} = x^{-\frac{1}{3}}$と変形します。

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

$= \frac{x^{\frac{5}{3}}}{\frac{5}{3}} - 2\frac{x^{\frac{2}{3}}}{\frac{2}{3}} + \log|x| + C = \frac{3}{5}x^{\frac{5}{3}} - 3x^{\frac{2}{3}} + \log|x| + C$

**(2) $\int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \frac{1-\cos^3x}{\cos^2x} dx$**

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

$\int_{\frac{\pi}{4}}^{\frac{\pi}{3}} (\sec^2x - \cos x) dx = [\tan x - \sin x]_{\frac{\pi}{4}}^{\frac{\pi}{3}}$

$= (\tan \frac{\pi}{3} - \sin \frac{\pi}{3}) - (\tan \frac{\pi}{4} - \sin \frac{\pi}{4}) = (\sqrt{3} - \frac{\sqrt{3}}{2}) - (1 - \frac{\sqrt{2}}{2}) = \frac{\sqrt{3}}{2} - 1 + \frac{\sqrt{2}}{2}$

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

$\cos^2x = 1 - \sin^2x = (1+\sin x)(1-\sin x)$

$\int \frac{(1+\sin x)(1-\sin x)}{1+\sin x} dx = \int (1-\sin x) dx = x + \cos x + C$

**(4) $\int_{0}^{1} (e^x + e^{-x})^2 dx$**

$(e^x + e^{-x})^2 = e^{2x} + 2 + e^{-2x}$

$\int_{0}^{1} (e^{2x} + 2 + e^{-2x}) dx = [\frac{1}{2}e^{2x} + 2x - \frac{1}{2}e^{-2x}]_{0}^{1}$

$= (\frac{1}{2}e^2 + 2 - \frac{1}{2}e^{-2}) - (\frac{1}{2} + 0 - \frac{1}{2}) = \frac{1}{2}e^2 + 2 - \frac{1}{2}e^{-2}$

**(5) $\int_{0}^{1} xe^{x^2+1} dx$**

$u = x^2+1$と置換すると、$du = 2x dx$。$x=0$のとき$u=1$、$x=1$のとき$u=2$。

$\int_{0}^{1} xe^{x^2+1} dx = \frac{1}{2} \int_{1}^{2} e^u du = \frac{1}{2} [e^u]_{1}^{2} = \frac{1}{2}(e^2 - e)$

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

$u = x^3-2x+1$と置換すると、$du = (3x^2-2) dx$。

$\int \frac{3x^2-2}{x^3-2x+1} dx = \int \frac{1}{u} du = \log|u| + C = \log|x^3-2x+1| + C$

**(7) $\int_{1}^{e} \sqrt{x} \log x dx$**

部分積分を用いる。$u = \log x$、$dv = \sqrt{x} dx$とすると、$du = \frac{1}{x} dx$、$v = \frac{2}{3}x^{\frac{3}{2}}$。

$\int_{1}^{e} \sqrt{x} \log x dx = [\frac{2}{3}x^{\frac{3}{2}} \log x]_{1}^{e} - \int_{1}^{e} \frac{2}{3}x^{\frac{3}{2}} \frac{1}{x} dx$

$= (\frac{2}{3}e^{\frac{3}{2}} \log e - \frac{2}{3}1^{\frac{3}{2}} \log 1) - \frac{2}{3}\int_{1}^{e} x^{\frac{1}{2}} dx = \frac{2}{3}e^{\frac{3}{2}} - \frac{2}{3}[\frac{2}{3}x^{\frac{3}{2}}]_{1}^{e}$

$= \frac{2}{3}e^{\frac{3}{2}} - \frac{4}{9}(e^{\frac{3}{2}} - 1) = \frac{2}{3}e^{\frac{3}{2}} - \frac{4}{9}e^{\frac{3}{2}} + \frac{4}{9} = \frac{2}{9}e^{\frac{3}{2}} + \frac{4}{9}$

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3. 最終的な答え

(1) $\frac{3}{5}x^{\frac{5}{3}} - 3x^{\frac{2}{3}} + \log|x| + C$

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

(3) $x + \cos x + C$

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

(5) $\frac{1}{2}(e^2 - e)$

(6) $\log|x^3-2x+1| + C$

(7) $\frac{2}{9}e^{\frac{3}{2}} + \frac{4}{9}$