Using partial fraction to power that less than 1

Question

I need help to check this answer since, unfortunately, the worksheet doesn't attach the key.

Integrate: $$\int \frac{1}{\sqrt{x} - \sqrt[3]{x}}dx$$

So, I tried use a substitution that allows the variables to become polynomials. I rewrote the expression to $$\int \frac{1}{x^\frac{5}{6}\left(x^\frac{-2}{6}- x^\frac{-3}{6}\right)} \, dx$$ when I set $$u = x^\frac{1}{6} \text{ and } \frac{dx}{x^\frac{5}{6}}=6 \, du$$ then $$\int \frac{6 \, du}{\frac{1}{u^2}-\frac{1}{u^3}} = \int \frac{6 u^3 \, du}{u-1}.$$ This becomes $$6 \int \left( u^2 + u + 1 + \frac{1}{u-1} \right) \, du = 6 \left(\frac{u^3}{3} + \frac{u^2}{2} + u + \ln\lvert u-1\rvert \right) + c.$$ and last part is to substitute back $u = x^\frac{1}{6}$.

I don't know whether this right or not. This is first time I am dealing with partial factions that have less than one power. Please correct it if this work is wrong. Thanks in advance.

Answer 1

This is correct. As a check: \begin{align*} \frac{\mathrm{d}}{\mathrm{d}u} & 6 \left( \frac{u^3}{3} + \frac{u^2}{2} + u + \ln |u-1| \right) \\ &= 6 \left( u^2 + u + 1 + \frac{1}{u-1} \right) \\ &= \frac{6u^3}{u-1} \text{,} \end{align*} as expected.

It can be a little easier to start with $\mathrm{lcm}(2,3) = 6$, so $$ \frac{1}{x^{1/2} - x^{1/3}} = \frac{1}{x^{3/6} - x^{2/6}} $$ and $u = x^{1/6}$, so both $x^{5/6} = u^5$ and $\mathrm{d}u = \frac{1}{6x^{5/6}} \,\mathrm{d}x$, and we conclude $\mathrm{d}x = 6u^5 \,\mathrm{d}u$. Substituting into the integral leaves $$ \int \frac{6 u^5 \,\mathrm{d}u}{u^3 - u^2} = \int \frac{6 u^3 \,\mathrm{d}u}{u - 1} \text{,} $$ and continue as you did.

Answer 2

An alternate view it to do a power expansion and the cancel powers. In this example $u^3$ can be seen as: $$u^3 = ( (u-1) + 1)^3 = (u-1)^3 + 3 \, (u-1)^2 + 3 \, (u-1) + 1$$ and $$\frac{u^3}{u-1} = (u-1)^2 + 3 \, (u-1) + 3 + \frac{1}{u-1}.$$ Integration yields: \begin{align} I &= \int \frac{u^3 \, du}{u-1} \\ &= \int (u-1)^2 \, du + 3 \, \int (u-1) \, du + 3 u + \int \frac{du}{u-1} \\ &= \frac{(u-1)^3}{3} + \frac{3 \, (u-1)^2}{2} + 3 \, u + \ln(u-1) + c_{0} \\ 6 I &= 2 (u-1)^3 + 9 (u-1)^2 + 18(u-1) + 6 \, \ln(u-1) + c_{1} \end{align}

This leads to the form $$\int \frac{dx}{\sqrt{x} - \sqrt[3]{x}} = 2 (\sqrt[6]{x}-1)^3 + 9 (\sqrt[6]{x}-1)^2 + 18(\sqrt[6]{x}-1) + 6 \, \ln(\sqrt[6]{x}-1) + c_{1}. $$