0
$\begingroup$

Find the integral $$\int_{1}^{2}\sqrt[3]{\frac{2-x}{x^7}}dx $$

Could not find a correct solution.

  • 0
    I tried to separate (2-x)/x^7 inside the root. But could not come up with something that would keep me going – 2017-01-17

3 Answers 3

4

\begin{align} I=\int_{1}^{2}\sqrt[3]{\frac{2-x}{x^7}}dx=\int_{1}^{2}\frac1{x^2}\sqrt[3]{\frac{2-x}{x}}dx \end{align}

Here you could then use $\frac{2-x}{x}=u^3$ which implies $x=2(1+u^3)^{-1}$ and $dx=-6u^2(1+u^3)^{-2}$ so that your integral becomes

\begin{align} I=\int_{0}^{1}\frac{(1+u^3)^2}{4}\times u\times 6u^2(1+u^3)^{-2} du \end{align}

  • 0
    Oh now it is much easier to solve. Thanks – 2017-01-17
  • 1
    you welcome ... in general you should almost alway get rid of roots if you can ... – 2017-01-17
  • 0
    @Math-fun: Are you sure of your coefficients? We globally have the same method, with a slightly different implementation, but I can't find my error. – 2017-01-17
1

Hint:

Set $u=\sqrt[3]{\dfrac{2-x}{x}}\iff u^3=\dfrac{2-x}{x}==\dfrac{2}{x}-1$.

Differentiating, we obtain $\quad3u^2\,\mathrm du=-\dfrac{2}{x^2}\,\mathrm d x,\enspace\text{whence}\quad\mathrm d x= -\dfrac32u^2x^2\,\mathrm du $, so $$\int_{1}^{2}\sqrt[3]{\frac{2-x}{x^7}}\,\mathrm d x=-\frac32\int_{1}^{0}\frac1{x^2}\,u\cdot u^2x^2\,\mathrm du=\frac32\int_{0}^{1}u^3\,\mathrm du.$$

  • 0
    Your answer provides twice the value :-) I could take the liberty of correcting, shall I? – 2017-01-17
  • 0
    @Math-fun: You could, of course, no problem, but I've just found I repeated a `2` on extracting $x^6$ from the cubic root. – 2017-01-17
1

$$I=\int_{1}^{2}\sqrt[3]{\frac{2-x}{x^7}}dx=\int_{1}^{2}\frac1{x^2}\sqrt[3]{\frac2x-1}dx$$

Now choose $\frac2x-1=u\implies-\dfrac2{x^2}dx=du$

$$I=-\dfrac12\int_1^0u^{-1/3}du=?$$