minikim38

2022-01-29

How to calculate the indefinite integral of $\frac{1}{{x}^{\frac{2}{3}}(1+{x}^{\frac{2}{3}})}$ ?

I substituted,

$t=\frac{1}{{x}^{\frac{1}{3}}}$

$\frac{dt}{dx}=-\frac{1}{3{x}^{\frac{4}{3}}}$

$\frac{dt}{dx}=-\frac{{t}^{4}}{3}$

Rewriting the question,

$\int \frac{dx}{{x}^{\frac{2}{3}}+{x}^{\frac{4}{3}}}$

$-\frac{1}{3}\int \frac{dt}{{t}^{4}(\frac{1}{{t}^{2}}+\frac{1}{{t}^{4}})}$

We have,

$-\frac{1}{3}\int \frac{dt}{{t}^{2}+1}$

$-\frac{1}{3}{\mathrm{tan}}^{-1}t+C$

$-\frac{1}{3}{\mathrm{tan}}^{-1}\left(\frac{1}{{x}^{\frac{1}{3}}}\right)+C$

But the answer given is

$3{\mathrm{tan}}^{-1}{x}^{\frac{1}{3}}+C$

Where am I wrong?

I substituted,

Rewriting the question,

We have,

But the answer given is

Where am I wrong?

Alfred Mueller

Beginner2022-01-30Added 10 answers

Two observations resolve this. The first is that you didn't invert the derivative properly, so your 1/3 coefficient should be 3 because $dx=-3{t}^{-4}dt$ . Secondly, $\frac{\mathrm{arctan}1}{y}=\frac{\pi}{2}-\mathrm{arctan}y$ implies there's more than one way to write the answer, with your method getting something valid. I think you were expected to substitute $t={x}^{\frac{1}{3}}$ instead; your choice does work, but it's conceptually more complex.

utgyrnr0

Beginner2022-01-31Added 11 answers