#1 懷念的微積分與 LaTeX
發表於 : 2021年 11月 21日, 16:28
版上有人問起,真是令人懷念啊……
連著練習一陣子沒用的\(\LaTeX\),試著計算於下:
\(\begin{align}
\int \frac{x^3+4x^2+x-1}{x^3+x^2} dx&= \int \frac{(x^3+x^2)+(3x^2+x-1)}{x^3+x^2}dx \\
&= \int \left( \frac{x^3+x^2}{x^3+x^2} + \frac{3x^2+x-1}{x^3+x^2} \right)dx \\
&= \int \left( 1 + \frac{3x^2+x-1}{x^3+x^2} \right)dx \\
&= \int 1 dx + \int \left( \frac{3x^2+x-1}{x^3+x^2} \right)dx \\
&= \int 1 dx + \bbox[5px, border: 2px solid green]{\int \left[ \frac{3x^2+x-1}{x^2(x+1)} \right]dx}
\end{align}\)
題目要求「部分分式積分法」 求解,將綠框部份挪出計算於下:
\(\begin{align}
\frac{3x^2+x-1}{x^3+x^2} &= \frac{a}{x} + \frac{b}{x^2} + \frac{c}{x+1} \\
&=\frac{a \cdot x(x+1)}{x \cdot x(x+1)} + \frac{b \cdot (x+1)}{x^2 \cdot (x+1)} + \frac{c \cdot x^2}{(x+1) \cdot x^2} \\
&=\frac{ax^2 + ax + bx +b + cx^2}{x^3+x^2} \\
&\downarrow \\
3x^2+x-1&=(a+c)x^2+(a+b)x+b \\
&\downarrow \\
\begin{cases}a+c&=3\\
a+b&=1\\
b&=-1 \end{cases} &\rightarrow
\begin{cases}a&=2\\
b&=-1\\
c&=1 \end{cases}\\
&\downarrow \\
\frac{3x^2+x-1}{x^3+x^2} &= \bbox[5px, border: 2px solid rebeccapurple]{\frac{2}{x} + \frac{-1}{x^2} + \frac{1}{x+1}} \end{align}
\)
將紫框部份帶回原始算式,得:
\(\begin{align}
\int \frac{x^3+4x^2+x-1}{x^3+x^2} dx&= \int 1 dx + \int \left( \frac{2}{x} + \frac{-1}{x^2} + \frac{1}{x+1} \right)dx\\
&= \int 1 dx + \int \frac{2}{x}dx + \int \frac{-1}{x^2} dx + \int \frac{1}{x+1} dx\\
&= x+ C + 2\ln{|x|} + \frac{1}{x} + \ln {|x+1|} \\
&=\bf{2\ln{|x|} + \ln {|x+1|} + x+ \frac{1}{x} + C} \end{align}\)
歡迎大家也在版上恣意揮灑數學之美。(?)
連著練習一陣子沒用的\(\LaTeX\),試著計算於下:
\(\begin{align}
\int \frac{x^3+4x^2+x-1}{x^3+x^2} dx&= \int \frac{(x^3+x^2)+(3x^2+x-1)}{x^3+x^2}dx \\
&= \int \left( \frac{x^3+x^2}{x^3+x^2} + \frac{3x^2+x-1}{x^3+x^2} \right)dx \\
&= \int \left( 1 + \frac{3x^2+x-1}{x^3+x^2} \right)dx \\
&= \int 1 dx + \int \left( \frac{3x^2+x-1}{x^3+x^2} \right)dx \\
&= \int 1 dx + \bbox[5px, border: 2px solid green]{\int \left[ \frac{3x^2+x-1}{x^2(x+1)} \right]dx}
\end{align}\)
題目要求
\(\begin{align}
\frac{3x^2+x-1}{x^3+x^2} &= \frac{a}{x} + \frac{b}{x^2} + \frac{c}{x+1} \\
&=\frac{a \cdot x(x+1)}{x \cdot x(x+1)} + \frac{b \cdot (x+1)}{x^2 \cdot (x+1)} + \frac{c \cdot x^2}{(x+1) \cdot x^2} \\
&=\frac{ax^2 + ax + bx +b + cx^2}{x^3+x^2} \\
&\downarrow \\
3x^2+x-1&=(a+c)x^2+(a+b)x+b \\
&\downarrow \\
\begin{cases}a+c&=3\\
a+b&=1\\
b&=-1 \end{cases} &\rightarrow
\begin{cases}a&=2\\
b&=-1\\
c&=1 \end{cases}\\
&\downarrow \\
\frac{3x^2+x-1}{x^3+x^2} &= \bbox[5px, border: 2px solid rebeccapurple]{\frac{2}{x} + \frac{-1}{x^2} + \frac{1}{x+1}} \end{align}
\)
將紫框部份帶回原始算式,得:
\(\begin{align}
\int \frac{x^3+4x^2+x-1}{x^3+x^2} dx&= \int 1 dx + \int \left( \frac{2}{x} + \frac{-1}{x^2} + \frac{1}{x+1} \right)dx\\
&= \int 1 dx + \int \frac{2}{x}dx + \int \frac{-1}{x^2} dx + \int \frac{1}{x+1} dx\\
&= x+ C + 2\ln{|x|} + \frac{1}{x} + \ln {|x+1|} \\
&=\bf{2\ln{|x|} + \ln {|x+1|} + x+ \frac{1}{x} + C} \end{align}\)
歡迎大家也在版上恣意揮灑數學之美。(?)