Computing $\displaystyle \int{ \frac{\sqrt[3]{x}+2\sqrt[4]{x}}{\sqrt{x}(\sqrt{x}+\sqrt[3]{x})^2}}\,dx$

Question

$$\int{\sqrt[3]{x}+2\sqrt[4]{x}\over \sqrt{x}(\sqrt{x}+\sqrt[3]{x})^2}dx$$ I think i need to make a substitution but am having difficulties because of the different roots. $$\int \left(\sqrt{x \over x+1}+1\right)^2 \sqrt[4]{x \over x+1}dx$$

Answer 1

If $y^4=\frac{x}{x+1}$ then $x=\frac{y^4}{1-y^4}$ and $dx=\left(\frac{y^4}{1-y^4}\right)'dy$, which is rational.

Then $$\begin{align}\int\left(\sqrt{x \over x+1}+1\right)^2 \sqrt[4]{x \over x+1}dx&=\int\left(y^2+1\right)^2\cdot y\cdot\left(\frac{y^4}{1-y^4}\right)'dy\end{align}$$

The integrand in the right-hand side is a rational function. We have an algorithm to compute these (partial fraction decomposition ...).

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This is a particular case of the following type of integrals:

$$\int R\left(x,\sqrt[n]{\frac{ax+b}{cx+d}}\right)dx$$ where $R(x,y)$ is a rational function. All of these can be rationalized by the change of variable $y^n=\frac{ax+b}{cx+d}$.

In your exercise

$$\int\left(\sqrt{x \over x+1}+1\right)^2 \sqrt[4]{x \over x+1}dx=\int\left[\left(\sqrt[4]{\frac{x}{x+1}}\right)^2+1\right]\sqrt[4]{\frac{x}{x+1}}dx$$

It has the form above for $R(x,y)=(y^2+1)^2\cdot y$, $n=4$, $a=1$, $b=0$, $c=1$, and $d=1$. This writing becomes apparent after you turn the radicals to a common denominator.

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The first exercise is also a particular case of this type of integrals.

We can write (turning the exponents to a common denominator)

$$\begin{align}\int \frac{\sqrt{x}+2\sqrt[4]{x}}{\sqrt{x}(\sqrt{x}+\sqrt[3]{x})^2}dx&=\int\frac{(\sqrt[12]{x})^6+2(\sqrt[12]{x})^3}{(\sqrt[12]{x})^6((\sqrt[12]{x})^6+(\sqrt[12]{x})^4)^2}dx\end{align}$$

We can take $R(x,y)=\frac{y^6+2y^3}{y^6(y^6+y^4)^2}$, $n=12$, $a=1$, $b=0$, $c=0$, and $d=1$.