Integration involving square root and negative power of $x$

Question

$$ f(x) = \int\sqrt{1+x^{-2/3}}\,\mathrm dx $$

What I have attempted: $$ \int\left(\frac{\sqrt{x^{2/3}}}{\sqrt{x^{2/3}}}\right)\sqrt{1+x^{-2/3}}\,\mathrm dx =\int\frac{\sqrt{x^{2/3}+1}}{x^{1/3}}\,\mathrm dx $$ Then let $$\tan u = x^{1/3}$$ so, $$ \int\frac{\sqrt{\tan^{2}{u}+1})(3\sec^{2}{u}\tan^{2}{u}\,\mathrm du)}{\tan u} =\int3\sec^{3}{u}\tan{u}\,\mathrm du $$ I come up with this answer $$\sec^{3}({\tan^{-1}{x^{1/3}}})+C$$

Now,I found out that my answer is correct only when $x$ is positive.

What things did I miss out? (I don't need a new approach.)

Answer 1

Thank you for your answer. But what I need is not a new approach, I want to know what mistakes I made.

You're making following transformation$$\int\sqrt{1+\frac{1}{x^{2/3}}}\,\mathrm dx=\int\frac{\sqrt{x^{2/3}+1}}{\sqrt{x^{2/3}}}\,\mathrm dx=\int\frac{\sqrt{x^{2/3}+1}}{x^{1/3}}\,\mathrm dx$$

But notice that

$$\sqrt{x^{2/3}}= x^{1/3}$$ is only valid for $x\ge0$

For example put $x=-8$ $$\sqrt{(-8)^{2/3}}= 2=8^{1/3} $$

Answer 2

HINT.

1) Rewrite:

$$\sqrt{1+\left(\frac{1}{x^{1/3}}\right)^2}$$

2) Put:

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

3) and apply Integration by parts

$$ \int u(x) v'(x) \, dx = u(x) v(x) - \int u'(x) v(x) \, dx $$

where $v'(t)=t\sqrt{1+t^2}$

Answer 3

Let $$I = \int\sqrt{1+x^{-2/3}}dx\,$$ Change variable $$x^{-2/3}=\sinh^2(y)$$ that is to say $$x=\text{csch}^3(y)$$ $$dx=-3 \coth (y) \text{csch}^3(y)$$ So, $$I=-3\int \coth ^2(y) \text{csch}^2(y)dy\,=\coth ^3(y)$$