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$$$\frac{f_{1} \tan^{2}{\left(f \right)}}{g}$$$ 关于$$$g$$$的积分
您的输入
求$$$\int \frac{f_{1} \tan^{2}{\left(f \right)}}{g}\, dg$$$。
解答
对 $$$c=f_{1} \tan^{2}{\left(f \right)}$$$ 和 $$$f{\left(g \right)} = \frac{1}{g}$$$ 应用常数倍法则 $$$\int c f{\left(g \right)}\, dg = c \int f{\left(g \right)}\, dg$$$：
$${\color{red}{\int{\frac{f_{1} \tan^{2}{\left(f \right)}}{g} d g}}} = {\color{red}{f_{1} \tan^{2}{\left(f \right)} \int{\frac{1}{g} d g}}}$$
$$$\frac{1}{g}$$$ 的积分为 $$$\int{\frac{1}{g} d g} = \ln{\left(\left|{g}\right| \right)}$$$:
$$f_{1} \tan^{2}{\left(f \right)} {\color{red}{\int{\frac{1}{g} d g}}} = f_{1} \tan^{2}{\left(f \right)} {\color{red}{\ln{\left(\left|{g}\right| \right)}}}$$
因此，
$$\int{\frac{f_{1} \tan^{2}{\left(f \right)}}{g} d g} = f_{1} \ln{\left(\left|{g}\right| \right)} \tan^{2}{\left(f \right)}$$
加上积分常数：
$$\int{\frac{f_{1} \tan^{2}{\left(f \right)}}{g} d g} = f_{1} \ln{\left(\left|{g}\right| \right)} \tan^{2}{\left(f \right)}+C$$
答案
$$$\int \frac{f_{1} \tan^{2}{\left(f \right)}}{g}\, dg = f_{1} \ln\left(\left|{g}\right|\right) \tan^{2}{\left(f \right)} + C$$$A