∫ c o t x d x = ∫ c o s ( x ) s i n ( x ) d x {\displaystyle \int cotx\;dx=\int {\frac {cos(x)}{sin(x)}}dx}
u = 3 a x + b x 3 d u = ( 3 a + 3 b x 2 ) d x 1 3 d u = ( a + b x 2 ) d x {\displaystyle {\begin{aligned}u&=3ax+bx^{3}\\[2ex]du&=(3a+3bx^{2})dx\\[2ex]{\frac {1}{3}}du&=(a+bx^{2})dx\\[2ex]\end{aligned}}}
∫ a + b x 2 3 a x + b x 3 d x = ∫ 1 3 a x + b x 3 ( a + b x 2 ) d x = ∫ 1 3 a x + b x 3 ( a + b x 2 d x ) = 1 3 ∫ 1 u ( d u ) = 1 3 ∫ u − 1 / 2 d u = 1 3 u 1 2 1 2 + C = 2 3 ( 3 a x + b x 3 ) 1 / 2 + C = 2 3 3 a x + b x 3 + C {\displaystyle {\begin{aligned}\int {\frac {a+bx^{2}}{\sqrt {3ax+bx^{3}}}}dx&=\int {\frac {1}{\sqrt {3ax+bx^{3}}}}(a+bx^{2})\;dx=\int {\frac {1}{\sqrt {3ax+bx^{3}}}}(a+bx^{2}\;dx)\ \\[2ex]&={\frac {1}{3}}\int {\frac {1}{\sqrt {u}}}(du)={\frac {1}{3}}\int u^{-1/2}du\\[2ex]&={\frac {1}{3}}{\frac {u^{\frac {1}{2}}}{\frac {1}{2}}}+C\\[2ex]&={\frac {2}{3}}(3ax+bx^{3})^{1/2}+C\\[2ex]&={\frac {2}{3}}{\sqrt {3ax+bx^{3}}}+C\end{aligned}}}
![{\displaystyle {\begin{aligned}u&=3ax+bx^{3}\\[2ex]du&=(3a+3bx^{2})dx\\[2ex]{\frac {1}{3}}du&=(a+bx^{2})dx\\[2ex]\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/02464f0e436c19b198f5cf8d28fa89ba4633df4c)
![{\displaystyle {\begin{aligned}\int {\frac {a+bx^{2}}{\sqrt {3ax+bx^{3}}}}dx&=\int {\frac {1}{\sqrt {3ax+bx^{3}}}}(a+bx^{2})\;dx=\int {\frac {1}{\sqrt {3ax+bx^{3}}}}(a+bx^{2}\;dx)\ \\[2ex]&={\frac {1}{3}}\int {\frac {1}{\sqrt {u}}}(du)={\frac {1}{3}}\int u^{-1/2}du\\[2ex]&={\frac {1}{3}}{\frac {u^{\frac {1}{2}}}{\frac {1}{2}}}+C\\[2ex]&={\frac {2}{3}}(3ax+bx^{3})^{1/2}+C\\[2ex]&={\frac {2}{3}}{\sqrt {3ax+bx^{3}}}+C\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/4e66b85158f8454bb2bcdebe81912ca4ef4a99a9)