5.5 The Substitution Rule/17: Difference between revisions
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&= \frac{1}{3}\int \frac{1}{\sqrt{u}}(du) = \frac{1}{3}\int u^{-1/2} du \\[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{1}{3}\frac{u^{\frac{1}{2}}}{\frac{1}{2}} + C \\[2ex] | ||
&= \frac{2}{3}(3ax+bx^3) + C | &= \frac{2}{3}(3ax+bx^3)^{1/2} + C | ||
&= \frac{2}{3}{\sqrt{3ax+bx^3}} + C | |||
\end{align} | \end{align} | ||
</math> | </math> | ||
Revision as of 16:08, 14 September 2022
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \int \frac{a+bx^2}{\sqrt{3ax+bx^3}}dx }
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \begin{align} u &= 3ax+bx^3 \\[2ex] du &= (3a+3bx^2)dx \\[2ex] \frac{1}{3}du &= (a+bx^2)dx \\[2ex] \end{align} }
Failed to parse (Conversion error. Server ("https://en.wikipedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\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&={\frac {2}{3}}{\sqrt {3ax+bx^{3}}}+C\end{aligned}}}