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 &= \int \frac{1}{\sqrt{3ax+bx^3}}(a+bx^2)\;dx = \int \frac{1}{\sqrt{3ax+bx^3}}(a+bx^2\;dx)\ \\[2ex] | \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] | ||
Latest revision as of 19:18, 20 September 2022
![{\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)
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} \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{align} }