微积分(A)随缘一题[19]

试求：\(\int \frac{dx}{(1+x^2)^2}\)

\[\begin{aligned} \int \frac{dx}{(1+x^2)^2} =& \int \frac{\sec^2t dt}{\sec^4t} \\ =&\int \cos^2 t dt \\ =&\frac{1}{4}\int \cos 2t \cdot d(2t)+\frac{1}{2}\int dt \\ =&\frac{1}{4}\sin 2t+\frac{1}{2}t+C \\ =&\frac{1}{4}\sin 2\arctan x+\frac{1}{2}\arctan x+C \\ =&\frac{1}{2} \left(\sin \arctan x\cos \arctan x+\arctan x \right) + C \\ =&\frac{1}{2}\left( x \frac{\cos ^2 \arctan x}{\sin^2 \arctan x+\cos^2 \arctan x}+\arctan x\right)+C \\ =&\frac{1}{2}\left(\frac{x}{x^2+1}+\arctan x \right)+C \end{aligned} \]

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生活小妙招：

\[\begin{aligned} \int \frac{dx}{1+x^2} =&\frac{x}{1+x^2}-\int x \frac{-2x}{(1+x^2)^2}dx \\ =&\frac{x}{1+x^2}+\int \frac{2(x^2+1)-2}{(x^2+1)^2}dx \\ =&\frac{x}{1+x^2}+2 \int \frac{dx}{1+x^2}-2\int \frac{dx}{(1+x^2)^2} \end{aligned} \]

移项：

\[\begin{aligned} \int \frac{dx}{(1+x^2)^2}= -\frac{1}{2}\frac{x}{1+x^2} + \frac{1}{2}\int \frac{dx}{1+x^2} \end{aligned}=\frac{1}{2} \left(\frac{x}{1+x^2}+\arctan x \right)+C \]

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实际上：

\[\begin{aligned} \int \frac{dx}{(1+x^a)^n} =&\frac{x}{(1+x^a)^n}-\int \frac{-nax^{a}}{(1+x^a)^{n+1}}dx \\ =&\frac{x}{(1+x^a)^n}+na \int \frac{x^a+1-1}{(1+x^a)^{n+1}}dx \\ =&\frac{x}{(1+x^a)^n}+na\int \frac{dx}{(1+x^a)^{n}}-na \int \frac{dx}{(1+x^a)^{n+1}} \\ \end{aligned} \]

所以：

\[\int \frac{dx}{(1+x^a)^{n+1}}=\frac{x}{na(1+x^a)^n}+\frac{na-1}{na}\int \frac{dx}{(1+x^a)^n} \]

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能双元嘛？

\[\begin{aligned} \int \frac{dx}{(1+x^2)^2} \xlongequal[p^2=1+q^2,pdp=qdq]{p=\sqrt{1+q^2},q=x}&\int \frac{dq}{p^4} \\ =&\int \frac{d\frac{q}{p}}{p} \\ =&\int \sqrt{1-\left(\frac{q}{p}\right)^2}d\frac{q}{p} \\ =&\frac{1}{2}\left(\frac{x}{\sqrt{1+x^2}}\sqrt{1-\frac{x^2}{1+x^2}}+\arcsin \frac{x}{\sqrt{1+x^2}}\right)+C \\ =&\frac{1}{2}\left( \frac{x}{1+x^2}+\arcsin \frac{x}{\sqrt{1+x^2}} \right) \end{aligned} \]

其中：

\[d\frac{q}{p}=\frac{pdq-qdp}{p^2}=\frac{p-\frac{q^2}{p}}{p^2}dq=\frac{p^2-q^2}{p^3}dq=\frac{dq}{p^3} \]