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I am finding the ingetral of:

$$\int{x+3\over (3-2x)^{2\over 3}}$$

So,

$u=3-2x$, $_{}$ $du=-2 dx$,$_{}$ and $x={1\over 2}(u-3)$

$$\int{{1\over 2}(u-3)+3 \over u^{3\over 2}}({-{1\over 2}})du \\ = {-{1\over 2}}\int {{u\over 2} +{3\over 2}\over u^{2\over 3}} du \\ = {-{1\over 2}}\int{u+3 \over 2u^{2\over 3}} du \\ ={-{1\over 2}}\int {1\over 2} \cdot{u+3 \over u^{2\over 3}} du \\ = {-{1\over 4}}\int{u+3 \over u^{2\over 3}} du \\ = {-{1\over 4}}\int{u \over u^{2 \over 3}}du {-{1\over 4}}\int{-3 \over u^{2\over 3}}du \\ = {-{1\over 4}}\int u^{-{1\over 3}} du {-{3\over 4}}\int u^{-{2\over 3}} du \\ = {-{1\over 4}}({u^{2\over 3} \over {2 \over 3}}) + {3\over 4} ({u^{1\over 3} \over {1 \over 3}}) \\ = {-{3\over 8}}(u^{2\over 3})-{9\over 4}(u^{1\over 3}) \\ = {-{3\over 8}}(3-2x)^{2\over 3} + {9\over 4}(3-2x)^{1\over 3}$$

It turns out that this is wrong. What did I do wrong?

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    $\begingroup$ Doesn't $x=0.5(3-u)$? $\endgroup$ – Noam Dolovich Jun 11 '16 at 17:38
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    $\begingroup$ One observation: if $u=3-2x$, $x=\frac{1}{2}(3-u)$. That is not what you have. $\endgroup$ – MathematicsStudent1122 Jun 11 '16 at 17:38
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    $\begingroup$ Very simple mistake I made there... Now I have to do everything again $\endgroup$ – didgocks Jun 11 '16 at 17:40
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$$\int{x+3\over (3-2x)^{2\over 3}}dx$$ $$u=(3-2x)\Rightarrow du=-2dx$$ $$=-\frac{1}{2}\int\frac{\frac{1}{2}(3-u)+3}{u^\frac{2}{3}}du$$ $$=-\frac{1}{2}\int-\frac{u-9}{2u^\frac{2}{3}}du$$ $$=-\frac{1}{2}\left(-\frac{1}{2}\left(\int\frac{u}{u^\frac{2}{3}}du-\int\frac{9}{u^\frac{2}{3}}du\right)\right)$$ $$\int\frac{u}{u^\frac{2}{3}}du\Rightarrow \frac{3u^\frac{4}{3}}{4},\int\frac{9}{u^\frac{2}{3}}du\Rightarrow 27\sqrt[3]{u}$$ $$=-\frac{1}{2}\left(-\frac{1}{2}\left(\frac{3(3-2x)^\frac{4}{3}}{4}-27\sqrt[3]{3-2x}\right)\right)+C$$ $$=\frac{1}{4}\left(\frac{3}{4}(3-2x)^\frac{4}{3}-27\sqrt[3]{3-2x}\right)+C$$

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$$3-2x=u^3$$ $$\int{x+3\over (3-2x)^{2\over 3}}dx=-\frac{3}{4}\int(9-u^3)du=\frac{3}{16}u^4-\frac{27}{4}u+c$$

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