5.5 The Substitution Rule/17: Difference between revisions

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<math>
<math>
\int \frac{a+bx^2}{\sqrt{3ax+bx^3}}\;dx
\int \frac{a+bx^2}{\sqrt{3ax+bx^3}}dx
</math>
</math>


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\begin{align}
\begin{align}


\int \frac{a+bx^2}{\sqrt{3ax+bx^3}}dx &= \int \frac{1}{\sqrt{u}}du = \int\left(\frac{1}{x}dx\right)\sin{(\ln{(x)})} \\[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]


&= \int (du)\sin{(u)} = \int \sin{(u)}du \\[2ex]
 
&= -\cos{(u)} + C \\[2ex]
&= \frac{1}{3}\int \frac{1}{\sqrt{u}}(du) = \frac{1}{3}\int u^{-1/2} du \\[2ex]
&= -\cos{(\ln{(x)})} + C
&= \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}
\end{align}
</math>
</math>

Latest revision as of 19:18, 20 September 2022