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

Question

$$\int \frac{dx}{x^{2/3}(1+x^{2/3})}.$$

I substituted,

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

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

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

Rewriting the question,

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

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

We have,

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

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

$$-\frac{1}{3}\tan^{-1}\Biggl(\frac{1}{x^{1/3}}\Biggr)+C$$

But the answer given is $$3\tan^{-1}x^{1/3}+C$$

Where am I wrong?

Any help would be appreciated.

Answer 1

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

Answer 2

If $$t=x^{-2/3}$$ then $$dt=\frac{-2}{3}x^{-5/3}dx$$

Answer 3

$t=\sqrt[3]{x},\qquad dt=\frac{1}{x^{2/3}}dx$ $$\int \frac{1}{x^{\frac{2}{3}}+x^{\frac{4}{3}}}dx=3\int \frac{1}{t^2+1}dt$$