5.5 The Substitution Rule/45: Difference between revisions

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\begin{align}
\begin{align}


\int_{}^{} \left(\frac {x}{\sqrt[4]{x+2}}\right)dx &=\int_{}^{} \left(\frac{u-2}{\sqrt[4]{u}}\right) \\[2ex]
\int_{}^{} \left(\frac {x}{\sqrt[4]{x+2}}\right)dx &=\int_{}^{} \left(\frac{u-2}{\sqrt[4]{u}}\right)du \\[2ex]
&=\int_{}^{} \left(\frac{u}{\sqrt[4](u)} - \frac{2}{\sqrt[4](u)}\right) \\[2ex]
&=\int_{}^{} \left(\frac{u}{\sqrt[4](u)} - \frac{2}{\sqrt[4](u)}\right)du \\[2ex]
&=\int_{}^{} \left(u^{\frac{3}{4}} - 2u^{-\frac{1}{u}} \right) \\[2ex]
&=\int_{}^{} \left(u^{\frac{3}{4}} - 2u^{-\frac{1}{u}} \right)du \\[2ex]
&= \frac{4}{7} u^{\frac{7}{4}} - 2(\frac{4}{3})u^{\frac{3}{4}} + c \\[2ex]
&= \frac{4}{7} u^{\frac{7}{4}} - 2(\frac{4}{3})u^{\frac{3}{4}} + c \\[2ex]
&= \frac{4}{7} (x+2)^{\frac{7}{4}} - \frac{8}{3} (x+2)^{\frac{3}{4}} +c \\[2ex]
&= \frac{4}{7} (x+2)^{\frac{7}{4}} - \frac{8}{3} (x+2)^{\frac{3}{4}} +c \\[2ex]

Latest revision as of 22:21, 7 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_{}^{} \left(\frac {x}{\sqrt[4]{x+2}}\right)dx }

Failed to parse (Conversion error. Server ("https://en.wikipedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\begin{aligned}\int _{}^{}\left({\frac {x}{\sqrt[{4}]{x+2}}}\right)dx&=\int _{}^{}\left({\frac {u-2}{\sqrt[{4}]{u}}}\right)du\\[2ex]&=\int _{}^{}\left({\frac {u}{{\sqrt[{4}]{(}}u)}}-{\frac {2}{{\sqrt[{4}]{(}}u)}}\right)du\\[2ex]&=\int _{}^{}\left(u^{\frac {3}{4}}-2u^{-{\frac {1}{u}}}\right)du\\[2ex]&={\frac {4}{7}}u^{\frac {7}{4}}-2({\frac {4}{3}})u^{\frac {3}{4}}+c\\[2ex]&={\frac {4}{7}}(x+2)^{\frac {7}{4}}-{\frac {8}{3}}(x+2)^{\frac {3}{4}}+c\\[2ex]\end{aligned}}}