5.5 The Substitution Rule/17: Difference between revisions
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\begin{align} | \begin{align} | ||
\int \frac{a+bx^2}{\sqrt{3ax+bx^3}}dx &= \frac{1}{3}\int \frac{1}{\sqrt{u}}du = | \int \frac{a+bx^2}{\sqrt{3ax+bx^3}}dx &= \frac{1}{3}\int \frac{1}{\sqrt{u}}du = \frac{1}{3}\int \{u^{-1/2}du [2ex] | ||
&= \int (du)\sin{(u)} = \int \sin{(u)}du \\[2ex] | &= \int (du)\sin{(u)} = \int \sin{(u)}du \\[2ex] | ||
Revision as of 23:10, 13 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&={\frac {1}{3}}\int {\frac {1}{\sqrt {u}}}du={\frac {1}{3}}\int \{u^{-1/2}du[2ex]&=\int (du)\sin {(u)}=\int \sin {(u)}du\\[2ex]&=-\cos {(u)}+C\\[2ex]&=-\cos {(\ln {(x)})}+C\end{aligned}}}