与えられた10個の不定積分を計算します。 (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) $\int \tanh x dx$

はい、承知いたしました。積分問題ですね。今回は、(1)から(10)の問題すべて解いてみます。

1. 問題の内容

与えられた10個の不定積分を計算します。

(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) $\int \tanh x dx$

2. 解き方の手順

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

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

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

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

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

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

$\int \frac{x}{x^2+4} dx = \frac{1}{2}\int \frac{2x}{x^2+4} dx = \frac{1}{2}\ln(x^2+4) + C_1$

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

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

$\frac{1}{4} \int \frac{1}{u^2+1} 2du = \frac{1}{2}\arctan(u) = \frac{1}{2}\arctan(\frac{x}{2}) + C_2$

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

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

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

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

$= \frac{1}{2} [-\frac{1}{8}\cos 8x - \frac{1}{2}\cos 2x] + C = -\frac{1}{16}\cos 8x - \frac{1}{4}\cos 2x + C$

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

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

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

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

$u = \cosh x$と置換すると、$du = \sinh x dx$

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

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