eduction
\[ \begin{align} 假设: F(u) 是以x为自变量的复合函数 \\ \quad F^{\prime}(u)=f(u) \\ \text { 设: } u=\varphi(x) \\ \Rightarrow \int f(u) d x=F^{\prime}(u)+C, \quad(式0.0.0) \\ \\ 根据链式法则: \\ F^{\prime} (u)=F^{\prime}(u) \cdot(u)^{\prime}=f(u) \cdot(u)^{\prime} \\ \Rightarrow \int f(u) \cdot(u)^{\prime} d x=F^{\prime}(x u)+C, \quad(式1.1 .1) \\ \quad \therefore(式0,0.0)=(式1.1 .1) \\ \\ \because(u)^{\prime} d x=d u \\ \Rightarrow \int f(u) d u=F^{\prime}(u)+C \\ \\ \int f[\varphi(x)] \cdot \varphi^{\prime}(x) d x=\int f[\varphi(x)] d[\varphi(x)] =F^{\prime}[\varphi(x)]+C \end{align} \]example 0
\[\begin{align} 已知: \int f(x) d x=x^{2}+C \\ \int x f\left(1-x^{2}\right) d x=? \\ \\ \\ \Rightarrow\int f\left(1-x^{2}\right) \cdot x d x \\ \\ \Rightarrow\int f\left(1-x^{2}\right) \cdot-\frac{1}{2}(-2 x d x) \\ \\ \because-2 x=[F(x)+ C]^{\prime} \\ \\ \therefore-2 x=\left(1-x^{2}\right)^{\prime}, \quad-2 x d x=d\left(1-x^{2}\right) \\ \\ \Rightarrow-\frac{1}{2} \int f\left(1-x^{2}\right) \cdot d\left(1-x^{2}\right) \\ \\ \because\left(1-x^{2}\right) \Leftrightarrow x \\ \\ \therefore-\frac{1}{2} \int f\left(1-x^{2}\right) \cdot d\left(1-x^{2}\right)=-\frac{1}{2}\left(1-x^{2}\right)^{2}+C \end{align} \]标签：prime,right,第一,换元,int,varphi,cdot,积分法,left From： https://www.cnblogs.com/Preparing/p/18188247