（A）

• 1.利用不定積分換元法則（I）計算下列不定積分

(1) \(\int \sin(\omega x+\psi)dx\)

(2) \(\int \frac{10}{\sqrt[3]{3-5x}}dx\)

(3) \(\int \frac{dx}{\sqrt{1-16x^2}}\)

(4) \(\int {x^2} (3+2x^2)^{\frac{1}{6}} dx\)
(5)\(\int \frac{3x^3+x}{1+x^4}dx\)
(6)\(\int \frac{\sqrt{1+\sqrt x}}{\sqrt x}dx\)

(7) \(\int\frac{\cos \ln\vert x\vert}{x}dx\)

(8) \(\int\frac{ln ln x}{x\ln x}dx (x>e)\)

(9) \(\int\frac{cos^3x}{sin^2x}dx\)
(10)\(\int cos^4xdx\)
(11)\(\int sin^2 cos^2xdx\)
(12)\(\int sec^4xdx\)
(13)\(\int csc^3xcotxdx\)
(14)\(\int \frac{dx}{e^x+1}\)
(15)\(\int \frac{dx}{1+sin^2x}dx\)
(16)\(\int \frac{x}{\sqrt{1+x^2}}e^{-\sqrt{1+x^2}}dx\)

解：

(1)
\(\begin{aligned} I&= \int \sin(\omega t+\psi)dx\\ &=-\frac{1}{\omega} cos{\omega x+\psi}+C \end{aligned}\)

(2)
\(\begin{aligned} I&=\int \frac{10}{\sqrt[3]{3-5x}}dx\\ &= -3(3-5x)^{\frac{2}{3}}+C \end{aligned}\)

(3)
\(\begin{aligned} I&=\int \frac{dx}{\sqrt{1-16x^2}} \\ &= \frac{1}{4}arcsin4x+C \end{aligned}\)

(4)

\(\begin{aligned} I&=\int {x^2} (3+2x^2)^{\frac{1}{6}} dx \\ &= \frac{1}{3}\int (3+2x^3)^{\frac{1}{6}}dx \\ &= \frac{1}{3}\times\frac{6}{7\times2}(3+2x^3)^{\frac{7}{6}}+C \\ &=\frac{1}{7}(3+2x^3)^{\frac{7}{6}}+C \end{aligned}\)

(5)

\(\begin{aligned} I&=\int \frac{3x^3+x}{1+x^4}dx \\ &=\int (\frac{3x^3}{1+x^4}+\frac{x}{1+x^4})d x \\ &= \int \frac{\frac{3}{4}}{1+x^4}dx^4+\int \frac{\frac{1}{2}}{1+x^4}dx^2 \\ &=\frac{3}{4}\ln(1+x^4)+\frac{1}{2}\arctan x+C \end{aligned}\)

(6)
\(\begin{aligned} I&=\int \frac{\sqrt{1+\sqrt x}}{\sqrt x}dx \\ &= 2\int \sqrt{1+\sqrt x}dx \\ &= \frac{4}{3}(1+\sqrt x)^{\frac{3}{2}}+C \end{aligned}\)

(7)
\(\begin{aligned} I&=\int\frac{\cos \ln\vert x\vert}{x}dx \\ &= \int \cos \ln|x|d\ln|x| \\ &=\sin \ln |x|+C \end{aligned}\)

(8)
\(\begin{aligned} I&=\int\frac{\cos \ln\vert x\vert}{x}dx \\ &= \int \frac{\ln \ln x}{\ln x}d\ln x \\ &=\int \ln \ln xd\ln \ln x \\ &=\frac{1}{2}(\ln \ln x)^2+C \end{aligned}\)

(9)
\(\begin{aligned} I&=\int\frac{\cos^3x}{\sin^2x}dx\\ &= \int \frac{1-\sin ^2x}{\sin^2x}d\sin x \\ &= \int \frac{1}{\sin^2x}d\sin x-\int 1d\sin x \\ &=-\frac{1}{\sin x}-\sin x+C \end{aligned}\)

(10)
\(\begin{aligned} I&=\int cos^4xdx \\ &= x\cos x-\int x(-4\cos^2 x\cos x\sin x)dx \\ &=x\cos x+\int x(1+ \cos 2x)sin2x\ dx \\ &=xcos^4x-\frac{1}{2}(\int xd\cos2x+\int x\cos2xd\cos2x)以上三步是我胡幾把寫的\\ &=\int(\frac{1+2cos2x}{2})^2dx\\ &=\frac{1}{4}\int(1+2\cos2x+\frac{1+cos4x}{2})dx\\ &=\frac{1}{32}(12x+8\sin2x+\sin 4x)+C \end{aligned}\)

(11)
\(\begin{aligned} I&=\int \sin^2 x\cos^2xdx\\ &= \int \frac{1}{4}\sin^22xdx \\ &=\int \frac{1}{4}\times\frac{1-\cos 4x}{2}dx \\ &=\frac{1}{8}(x-\int\cos4xdx)\\ &=\frac{1}{32}(4x-\sin 4x)+C \end{aligned}\)

(12)
\(\begin{aligned} I&=\int sec^4xdx \\ &= \int \sec^2xd\tan x \\ &= \int (1+\tan^2x)d\tan x \\ &=\tan x+\frac{1}{3}\tan^3x+C \end{aligned}\)

(13)
\(\begin{aligned} I&=\int csc^3xcotxdx \\ &=\int \frac{\cos x}{\sin^4x}dx \\ &= \int \frac{1}{\sin^4x}d\sin x \\ &=-\frac{1}{3}\sin^{-3}xdx \end{aligned}\)

(14)
\(\begin{aligned} I&=\int \frac{dx}{e^x+1} \\ &= \int \frac{e^x}{e^x(e^x+1)}dx \\ &= \int (\frac{1}{e^x}-\frac{1}{e^x+1})dx \\ &=x-\int \frac{1}{e^x+1}dx\\ &=x-ln(e^x+1)+C \end{aligned}\)
F2:分子分母同乘\(e^x\)
F3:分子化成\(e^x+1-e^x\)

(15)
\(\begin{aligned} I&=\int \frac{dx}{1+sin^2x} \\ &=\int \frac{2}{3-\cos 2x}dx \\ &= \int \frac{1+\tan x}{2+\tan^2 x}dx \\ &\overset{t=\tan x}{===}\int\frac{1+t^2}{2+t^2}d\arctan t\\ &=\frac{1}{\sqrt{2}}\arctan (\sqrt 2\tan x )+C \end{aligned}\)

(16)
\(\begin{aligned} I&=\int \frac{x}{\sqrt{1+x^2}}e^{-\sqrt{1+x^2}}dx\\ &=-e^{-\sqrt{1+x^2}} \\ &= \\ &= \end{aligned}\)

(17)
\(\begin{aligned} I&=\int \frac{\sqrt{\arctan x}}{1+x^2}dx \\ &=\int \sqrt{\arctan x}d\arctan x \\ &=\frac{2}{3}(\arctan x)^{\frac{3}{2}}+C \\ &= \end{aligned}\)

(18)
\(\begin{aligned} I&=\int \frac{dx}{\sqrt{1-x^2}\arccos {\frac{x}{2}}}dx\\ &=\int \frac{-1}{\arctan x}d\arctan x \\ &=-\ln |\arccos \frac{x}{2}|+C \end{aligned}\)

(19)
\(\begin{aligned} I&=\int tan^3xsecxdx \\ &=\int\frac{1-\cos^2x}{\cos^4x}d\cos x \\ &= \frac{1}{3}\sec^3x-\sec x+C \end{aligned}\)

(20)
\(\begin{aligned} I&=\int \frac{dx}{x^2-2x-3}\\ &=\int \frac{1}{(x-1)^2+2}dx \\ &= \frac{1}{\sqrt 2}\int \frac{1}{(\frac{x-1}{\sqrt 2})^2+1}d_{\frac{x-1}{\sqrt 2}} \\ &=\frac{1}{\sqrt 2}\arctan (\frac{x-1}{\sqrt 2})+C \end{aligned}\)

(21)
\(\begin{aligned} I&=\int \frac{dx}{\sqrt{1+x-x^2}}\\ &=\int \frac{1}{\sqrt{\frac{5}{4}-(x-\frac{1}{2})^2}} dx \\ &= \int \frac{1}{\frac{\sqrt{5}}{2}\sqrt{1-(\frac{2x-1}{\sqrt{5}})^2}} dx \\ &=\arcsin (\frac{2x-1}{\sqrt{5}})+C \end{aligned}\)

(22)
\(\begin{aligned} I&=\int \frac{sinxcos}{1-sin^4}dx\\ &=\int \frac{\sin x}{1-\sin^4x}d\sin x \\ &=\frac{1}{4}\int(\frac{1}{1-\sin^2x}+\frac{1}{1-sin^2x})d\sin^2x \\ &=\frac{1}{4}\ln \frac{1+sin^2x}{1-sin^2x}+C \end{aligned}\)

(23)
\(\begin{aligned} I&=\int \frac{sinxcos}{\sqrt[5]{sinx-cosx}}dx\\ &= \int \frac{1}{\sqrt[5]{\sin x-\cos x}}d(\sin x-\cos x) \\ &=\frac{5}{4}(\sin x-\cos x)^{\frac{4}{5}}+C \end{aligned}\)

(24)
\(\begin{aligned} I&=\int \frac{dx}{e^x+e^{\frac{x}{2}}}\\ &=-2\int \frac{e^{-\frac{x}{2}}}{1+e^{-\frac{x}{2}}}de^{-\frac{x}{2}} \\ &=-2e^{-\frac{x}{2}} +2\ln(1+e^{-\frac{x}{2}} )+C \\ &=2\ln \frac{e^{\frac{x}{2}}+1}{e^{\frac{x}{2}}}-e^{-\frac{x}{2}}+C'這兩個結果都對，只相差一個常數 \end{aligned}\)

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5500+個字元，我也是沒誰了。。。。