How to Solve an Integral with Fractions and Long Division?

[tmt1]

I have this integral:

$$\int_{}^{} \frac{1}{\sqrt{x} - {x}^{\frac{1}{3}}} \,dx$$

So, I can set $u^6 = x$ then $6u^5 du = dx$.

I can substitute that in and get:

$$6 \int_{}^{} \frac {u^5}{u^3 - u^2} \,du$$

Then I can simplify to

$$6 \int_{}^{} \frac {u^3}{u - 1} \,du$$

But I'm unsure how to proceed from here.

 

[Prove It]

I have this integral:

$$\int_{}^{} \frac{1}{\sqrt{x} - {x}^{\frac{1}{3}}} \,dx$$

So, I can set $u^6 = x$ then $6u^5 du = dx$.

I can substitute that in and get:

$$6 \int_{}^{} \frac {u^5}{u^3 - u^2} \,du$$

Then I can simplify to

$$6 \int_{}^{} \frac {u^3}{u - 1} \,du$$

But I'm unsure how to proceed from here.

I would substitute $\displaystyle \begin{align*} v = u - 1 \implies \mathrm{d}v = \mathrm{d}u \end{align*}$ and the integral becomes

$\displaystyle \begin{align*} 6\int{\frac{u^3}{u - 1 }\,\mathrm{d}u} &= 6\int{ \frac{\left( v + 1 \right) ^3}{v}\,\mathrm{d}v } \end{align*}$

Can you proceed?

 

[tmt1]

I would substitute $\displaystyle \begin{align*} v = u - 1 \implies \mathrm{d}v = \mathrm{d}u \end{align*}$ and the integral becomes

$\displaystyle \begin{align*} 6\int{\frac{u^3}{u - 1 }\,\mathrm{d}u} &= 6\int{ \frac{\left( v + 1 \right) ^3}{v}\,\mathrm{d}v } \end{align*}$

Can you proceed?

So then I have to multiply out the cube, and get:

$$6 \int_{}^{} v^3 \,dv + 18 \int_{}^{} v^2 \,dv + 18 \int_{}^{} v \,dv + 6 \int_{}^{} \,dv $$

And then integrate from here. Is that the idea?

 

[Prove It]

So then I have to multiply out the cube, and get:

$$6 \int_{}^{} v^3 \,dv + 18 \int_{}^{} v^2 \,dv + 18 \int_{}^{} v \,dv + 6 \int_{}^{} \,dv $$

And then integrate from here. Is that the idea?

Don't forget that every term also needs to be divided by v...

 

[Greg]

\(\displaystyle \int\dfrac{u^3}{u-1}\,du=\int\dfrac{u^3-1+1}{u-1}\,du\)

\(\displaystyle =\int\dfrac{(u-1)(u^2+u+1)}{u-1}+\dfrac{1}{u-1}\,du\)

...

 

[soroban]

I have this integral:

$$6 \int \frac {u^3}{u - 1} \,du$$

But I'm unsure how to proceed from here.

How about Long Division?

. . . $$\frac{6u^3}{u-1} \;=\;6\left(u^2 + u + 1 + \frac{1}{u-1}\right)$$