はい、承知いたしました。画像にある不定積分の問題を解きます。

はい、承知いたしました。画像にある不定積分の問題を解きます。

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1. 問題の内容**

次の10個の不定積分を求めます。

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

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

(3) $\int (\sqrt{x} + \sqrt[3]{x}) dx$

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

(5) $\int (\sin x + \cos x)^2 dx$

(6) $\int \sin 5x \cos 3x dx$

(7) $\int \tan^2 x dx$

(8) $\int \sinh x dx$

(9) $\int \cosh x dx$

(10) $\int \tanh x dx$

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2. 解き方の手順**

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

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

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

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

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

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

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

(3) $\int (\sqrt{x} + \sqrt[3]{x}) dx$

$\int (\sqrt{x} + \sqrt[3]{x}) dx = \int (x^{1/2} + x^{1/3}) dx = \frac{x^{3/2}}{3/2} + \frac{x^{4/3}}{4/3} + C = \frac{2}{3}x^{3/2} + \frac{3}{4}x^{4/3} + C$

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

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

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

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

$u = \frac{x}{2}, du = \frac{1}{2} dx \Rightarrow dx = 2 du$

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

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

(5) $\int (\sin x + \cos x)^2 dx$

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

$\int (1 + \sin 2x) dx = x - \frac{1}{2} \cos 2x + C$

(6) $\int \sin 5x \cos 3x dx$

$\sin A \cos B = \frac{1}{2}[\sin(A+B) + \sin(A-B)]$

$\sin 5x \cos 3x = \frac{1}{2}[\sin(8x) + \sin(2x)]$

$\int \sin 5x \cos 3x dx = \frac{1}{2} \int (\sin 8x + \sin 2x) dx = \frac{1}{2} (-\frac{1}{8} \cos 8x - \frac{1}{2} \cos 2x) + C = -\frac{1}{16} \cos 8x - \frac{1}{4} \cos 2x + C$

(7) $\int \tan^2 x dx$

$\tan^2 x = \sec^2 x - 1$

$\int \tan^2 x dx = \int (\sec^2 x - 1) dx = \tan x - x + C$

(8) $\int \sinh x dx = \cosh x + C$

(9) $\int \cosh x dx = \sinh x + C$

(10) $\int \tanh x dx$

$\tanh x = \frac{\sinh x}{\cosh x}$

$\int \tanh x dx = \int \frac{\sinh x}{\cosh x} dx$

$u = \cosh x, du = \sinh x dx$

$\int \frac{1}{u} du = \ln|u| + C = \ln|\cosh x| + C$

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3. 最終的な答え**

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

(2) $\ln|x| + 4\sqrt{x} + x + C$

(3) $\frac{2}{3}x^{3/2} + \frac{3}{4}x^{4/3} + C$

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

(5) $x - \frac{1}{2} \cos 2x + C$

(6) $-\frac{1}{16} \cos 8x - \frac{1}{4} \cos 2x + C$

(7) $\tan x - x + C$

(8) $\cosh x + C$

(9) $\sinh x + C$

(10) $\ln|\cosh x| + C$