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I want to solve this integral and think about doing the following steps:

$1)\quad t=x^{1/3}$

$2)\quad x=t^3$

$3)\quad dx=2t^3\,dt$

How I can show $\sqrt{x}$ as $t$?

$$\frac{x^{1/2}+3}{2+x^{1/3}}$$ Thanks!

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Step 3) should really read $dx = 3t^2dt$ (if you want to continue with this substitution). – baudolino May 7 '13 at 17:05
Ahhh, didnt see it :) – Ofir Attia May 7 '13 at 17:12
up vote 4 down vote accepted

The substitution $t=x^{\frac{1}{6}}$ is better. Then $$\sqrt{x}=t^3,\;\;\sqrt[3]{x}=t^2 \\ dx=6t^5 dt,$$ so $$ \int{\frac{\sqrt{x}+3}{2+{\sqrt[3]{x}}}\ dx}=6\int{\frac{(t^3+3)t^5}{2+t^2}\ dt}. $$

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I think is more complicated no? I need to seperate them now? – Ofir Attia May 7 '13 at 17:36
You can use long division. – M. Strochyk May 7 '13 at 17:47
After I`m using long division I get this polynom ( after doing integration ) How do I get t of x now? thanks! $\frac{t^7}{7}-\frac{2t^5}{5}+\frac{3t^4}{4}+\frac{4t^3}{3}+\frac{6t^2}{2}-8t+ \frac{16}{(\sqrt{2})}arctg\frac{t}{\sqrt{2}}+6ln|t^2+2|+C$ – Ofir Attia May 8 '13 at 6:37
Ok, Got it! thanks! – Ofir Attia May 8 '13 at 6:52

If you want to use your substitution $t=x^{1/3}$: $$ \begin{equation*} I=\int \frac{x^{1/2}+3}{2+x^{1/3}}dx=3\int \frac{t^{2}( t^{3/2}+3) }{2+t}\,dt,\tag{1} \end{equation*} $$ then you could use the additional substitution $u=t^{1/2}=x^{1/6}$ to obtain $$ \begin{equation*} I=6\int \frac{u^{8}+3u^{5}}{u^{2}+2}\,du.\tag{2} \end{equation*} $$ To evaluate this integral rewrite the integrand as $$\frac{u^{8}+3u^{5}}{u^{2}+2}=u^{6}-2u^{4}+3u^{3}+4u^{2}-6u-8+\frac{12u+16}{u^{2}+2},$$ using polynomial long division.

However, in general when the integrand is of the form $$f(x)=g(x,x^{p/q},x^{r/s},\ldots ),$$ with $p,q,r,s,\ldots\in \mathbb{N}$, the standard substitution is $x=t^{k}$, where $k=\operatorname{lcm}(q,s,\ldots )$. In the present case as pointed out by M. Strochyk the direct substitution is thus $x=t^{6}$, which yields the same integral as $(2)$: $$ \begin{equation*} I=\int \frac{x^{1/2}+3}{2+x^{1/3}}dx=6\int \frac{t^{8}+3t^{5}}{t^{2}+2} \,dt,\qquad x=t^{6}\tag{3}. \end{equation*} $$

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I was about to point out the $\rm lcm$ thingy in the other answer. (+1) – Pedro Tamaroff May 7 '13 at 22:14

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