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 {{SandboxNote}}
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+<math>
+.\int \frac{\mathrm{d}x}{1+e^{x}}
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+\\
-{{文字环绕/上下型环绕|[[File:0755.PNG|thumb|center|none]]}}
+解：令1+e^{x}=t
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+\\
+则x=\ln \left ( t-1\right )
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+\\
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+\mathrm{d}x=\mathrm{d}\left [\ln \left ( t-1\right ) \right ]=\frac{1}{t-1}\mathrm{d}t
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+\\
-adrs
+原式=\int \frac{\mathrm{d}t}{t\left ( t-1\right )}
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+\\
-savoia
+=\int \left ( \frac{1}{t-1}-\frac{1}{t}\right )\mathrm{d}t
+\\
+=\int \frac{1}{t-1}\mathrm{d}t-\int \frac{1}{t}\mathrm{d}t
+\\
+=\ln \left | t-1\right |-\ln \left | t\right |+C
+\\
+=x-\ln \left ( 1+e^{x}\right )+C
+\\
+.\int \frac{x}{1+x^{4}}\mathrm{d}x
+\\
+解：原式=\int \frac{\frac{1}{2}\mathrm{d}x^{2}}{1+x^{4}}
+\\
+=\frac{1}{2}\arctan x^{2}+C
+\\
+.\int \frac{\ln{x}}{x}\mathrm{d}x
+\\
+解：原式=\int \ln x\mathrm{d}\left ( \ln x\right )=\frac{1}{2}\ln ^{2}x+C
+\\
+.\int \frac{1}{x\left ( a\ln x+b\right )}\mathrm{d}x
+\\
+解：原式=\frac{1}{a}\int \frac{\mathrm{d}\left ( a\ln x+b\right )}{a\ln x+b}=\frac{1}{a}\ln \left | a\ln x+b\right |+C
+\\
+.\int \frac{e^{x}}{1+e^{2x}}\mathrm{d}x
+\\
+解：原式=\int \frac{\mathrm{d}e^{x}}{1+e^{2x}}=\arctan e^{x}+C
+\\
+.\int \frac{e^{-x}}{\sqrt{1-e^{-2x}}}\mathrm{d}x
+\\
+解：原式=-\int \frac{\mathrm{d}e^{-x}}{\sqrt{1-e^{-2x}}}=-\arcsin e^{-x}+C
+\\
+.\int \sin^{n} x\cos x\mathrm{d}x
+\\
+解：原式=\int \sin^{n} x\mathrm{d}\left ( \sin x\right )
+\\
+\frac{\sin^{n+1} x}{n+1}+C
+\\
+.\int \cos ^{3}x\mathrm{d}x
+\\
+解：原式=\int \cos ^{2}x\mathrm{d}\left ( \sin x\right )
+\\
+=\int \left ( 1-\sin ^{2}x\right )\mathrm{d}\left ( \sin x\right )
+\\
+=\int \mathrm{d}\sin x-\int \sin ^{2}x\mathrm{d}\left ( \sin x\right )
+\\
+=\sin x-\frac{1}{3}\sin ^{3}x+C
+\\
+.\int \sec ^{3}x\tan x\mathrm{d}x
+\\
+解：原式=\int \frac{\sin x}{\cos ^{4}x}\mathrm{d}x
+\\
+=-\int \frac{\mathrm{d}\cos x}{\cos ^{4}x}
+\\
+=\frac{1}{3\cos ^{3}x}+C
+\\
+.\int \frac{\left ( \arcsin x+2\right )^{2}}{\sqrt{1-x^{2}}}\mathrm{d}x
+\\
+解：原式=\int \left ( \arcsin x+2\right )^{2}\mathrm{d}\left ( \arcsin x+2\right )=\frac{\left ( \arcsin x+2\right )^{3}}{3}+C
+\\
+.\int \frac{\sqrt{\arctan x+1}}{1+x^{2}}\mathrm{d}x
+\\
+解：原式=\int \sqrt{\arctan x+1}\mathrm{d}\left ( \arctan x+1\right )
+\\
+=\frac{2}{3}\left ( \arctan x+1\right )^{\frac{3}{2}}+C
+\\
+=\frac{2\left ( \arctan x+1\right )\sqrt{\arctan x+1}}{3}+C
+</math>
+----
+<math>
+.(17).\int \frac{\mathrm{d}x}{\sqrt{4-9x^{2}}}
+\\
+解：令\frac{3}{2}x=\sin t，
+\\
+则x=\frac{2}{3}\sin t
+\\
+\mathrm{d}x=\mathrm{d}\left ( \frac{2}{3}\sin t\right )=\frac{2}{3}\cos t\mathrm{d}t
+\\
+原式=\frac{1}{2}\cdot \frac{2}{3}\int \frac{\cos t\mathrm{d}t}{\sqrt{1-\sin^{2}t}}
+\\
+=\frac{1}{3}\int \mathrm{d}t=\frac{1}{3}t+C=\frac{1}{3}\arcsin \frac{3}{2}x+C
+\\
+(25).\int e^{\sin x}\cos x\mathrm{d}x
+\\
+解：原式=\int e^{\sin x}\mathrm{d}\left ( \sin x \right )
+\\
+=e^{\sin x}+C
+\\
+(26).\int e^{x}\cos e^{x}\mathrm{d}x
+\\
+解：原式=\int \cos e^{x}\mathrm{d}e^{x}
+\\
+=\sin e^{x}+C
+\\
+(34).\int \frac{\mathrm{d}x}{e^{x}-1}
+\\
+解：令e^{x}-1=t，则x=\ln \left ( t+1\right )，
+\\
+\mathrm{d}x=\frac{1}{t+1}\mathrm{d}t
+\\
+则原式=\int \frac{1}{t\left ( t+1\right )}\mathrm{d}t
+\\
+=\int \frac{1}{t}\mathrm{d}t-\int \frac{1}{t+1}\mathrm{d}t
+\\
+=\ln \left | t\right |-\ln \left | t+1\right |+C
+\\
+=\ln \left | e^{x}-1\right |+x+C
+\\
+.(1).\int x\sqrt{x+1}\mathrm{d}x
+\\
+解：令t=\sqrt{x+1}，则x=t^{2}-1，
+\\
+\mathrm{d}x=\mathrm{d}\left ( t^{2}-1\right )=2t\mathrm{d}t，
+\\
+原式=2\int t^{2}\left ( t^{2}-1\right )\mathrm{d}t
+\\
+=\frac{2}{5}t^{5}-\frac{2}{3}t^{3}+C
+\\
+=\frac{\left ( x+1\right )\left ( 6x-4\right )}{15}\sqrt{x+1}+C
+\\
+(9).\int \frac{1}{\sqrt[3]{x+1}+1}\mathrm{d}x
+\\
+解：令t=\sqrt[3]{x+1}，则x=t^{3}-1，
+\\
+\mathrm{d}x=\mathrm{d}\left (\sqrt[3]{x+1}+1\right )=3t^{2}\mathrm{d}t，
+\\
+原式=\frac{3t^{2}}{t+1}\mathrm{d}t
+\\
+=\int \frac{3t^{2}+3t-3t-3+3}{t+1}\mathrm{d}t
+\\
+=\int 3t\mathrm{d}t-\int 3\mathrm{d}t+\int \frac{3}{t+1}\mathrm{d}t
+\\
+=\frac{3}{2}t^{2}-3t+3\ln\left | t+1 \right |+C
+\\
+=\frac{3}{2}\sqrt[3]{\left ( x+1\right )^{2}}-3\sqrt[3]{x+1}+3\ln \left | \sqrt[3]{x+1}+1\right |+C
+\\
+(12).\int \frac{1}{\left ( 1+x^{2}\right )^{2}}\mathrm{d}x
+\\
+解：令t=\arctan x，则x=\tan t，
+\\
+\frac{1}{ 1+x^{2} }\mathrm{d}x=\mathrm{d}\tan x=\mathrm{d}t，
+\\
+原式=\int \frac{1}{1+\tan ^{2}t}\mathrm{d}t
+\\
+=\int \cos ^{2}t\mathrm{d}t
+\\
+=\frac{1}{2}\int \left ( \cos 2t+1\right )\mathrm{d}t
+\\
+=\frac{1}{4}\sin 2t+\frac{1}{2}\mathrm{d}t+C
+\\
+=\frac{1}{4}\cdot \frac{2x}{x^{2}+1}+\frac{1}{2}\arctan x+C
+\\
+=\frac{x}{2\left ( x^{2}+1\right )}+\frac{1}{2}\arctan x+C
+\\
+(14).\int \frac{1}{x\sqrt{x^{2}-1}}\mathrm{d}x
+\\
+解：令x=\frac{1}{\cos t}，则t=\arccos \frac{1}{x}，
+\\
+\mathrm{d}x=d\left ( \frac{1}{\cos t}\right )=\frac{\sin t}{\cos^{2}t}\mathrm{d}t，
+\\
+原式=\int \cos t \cdot \frac{\cos t}{\sin t}\cdot \frac{\sin t}{\cos^{2} t}\mathrm{d}t
+\\
+=\int  \mathrm{d}t
+\\
+=t+C
+\\
+=\arccos\frac{1}{x}+C
+\\
+(15).\int \frac{x^{2}}{\sqrt{1-x^{2}}}\mathrm{d}x
+\\
+解：令x=\sin t，则t=\arcsin x，
+\\
+\mathrm{d}x=\cos t\mathrm{d}t，
+\\
+原式=\int\frac{\sin^{2}t}{\cos t}\cdot \cos t\mathrm{d}t
+\\
+=\int \sin^{2}t\mathrm{d}t
+\\
+=\frac{1}{2}\int\left ( \cos 2t+1\right )\mathrm{d}t
+\\
+=\frac{1}{4}\sin 2t+\frac{1}{2}t+C
+\\
+=\frac{1}{2}x\sqrt{1-x^{2}}+\frac{1}{2}\arcsin x+C
+\\
+.(2).\int x\sin x\mathrm{d}x
+\\
+解：原式=-\int x\mathrm{d}\left ( \cos x\right )
+\\
+=-x\cos x+\int \cos x \mathrm{d}x
+\\
+=\sin x-x\cos x+C
+\\
+(4).\int\ln\left ( x^{2}+1\right )\mathrm{d}x
+\\
+解：原式=x\ln\left ( x^{2}+1\right )-\int x\mathrm{d}\left [ \ln\left ( x^{2}+1\right )\right ]
+\\
+=x\ln\left ( x^{2}+1\right )-2\int \frac{x^{2}}{x^{2}+1}\mathrm{d}x
+\\
+=x\ln\left ( x^{2}+1\right )-2\int \mathrm{d}x+2\int \frac{1}{x^{2}+1}\mathrm{d}x
+\\
+=x\ln\left ( x^{2}+1\right )-2x+2\arctan x+C
+\\
+(4).\int x^{2}e^{-x}\mathrm{d}x
+\\
+解：原式=-\int x^{2}\mathrm{d}e^{-x}
+\\
+=-\left (x^{2}e^{-x}-\int 2xe^{-x}\mathrm{d}x \right )
+\\
+=-x^{2}e^{-x}-2\int x\mathrm{d}e^{-x}
+\\
+=-x^{2}e^{-x}-2\left ( xe^{-x}-\int e^{-x}\mathrm{d}x\right )
+\\
+=-x^{2}e^{-x}-2 xe^{-x}+2\int e^{-x}\mathrm{d}x
+\\
+=-x^{2}e^{-x}-2 xe^{-x}-2e^{-x}
+\\
+=-\left ( x^2+2x+2\right )e^{-x}
+</math>

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