The integral $$\int \frac{dx}{x^2(x^4+1)^{3/4}}$$ equals to

We have to evaluate the indefinite integral

$$I=\int \frac{dx}{x^{2}\,(x^{4}+1)^{3/4}}.$$

At first sight the powers of $$x$$ in the denominator make a direct substitution difficult, so we try to remove the factor $$x^{-2}$$ by the standard trick $$x=\frac1t$$ (or, equivalently, $$t=\frac1x$$). This inversion often converts a negative power of $$x$$ into a positive power of the new variable.

Put

$$x=\frac{1}{t}\qquad\Longrightarrow\qquad dx=-\frac{1}{t^{2}}\,dt.$$

We also express the remaining factors in terms of $$t$$:

$$x^{2}=\frac{1}{t^{2}},\qquad x^{4}=\frac{1}{t^{4}}.$$

Substituting these four expressions into the integral, we get

$$I=\int \frac{\displaystyle -\frac{1}{t^{2}}\,dt}{\displaystyle\frac{1}{t^{2}}\left(\frac{1}{t^{4}}+1\right)^{3/4}}.$$

The factor $$\frac{1}{t^{2}}$$ in the numerator and denominator cancels immediately, because

$$\frac{-\dfrac{1}{t^{2}}}{\dfrac{1}{t^{2}}}=-1.$$ Hence

$$I=-\int\frac{dt}{\left(\dfrac{1}{t^{4}}+1\right)^{3/4}}.$$ To simplify the remaining parenthesis we write

$$\frac{1}{t^{4}}+1=\frac{1+t^{4}}{t^{4}} \quad\Longrightarrow\quad \left(\frac{1}{t^{4}}+1\right)^{3/4}=\left(\frac{1+t^{4}}{t^{4}}\right)^{3/4} =\frac{(1+t^{4})^{3/4}}{t^{3}}.$$

Substituting this back, the integrand becomes

$$\frac{1}{\left(\dfrac{1+t^{4}}{t^{4}}\right)^{3/4}} =\frac{t^{3}}{(1+t^{4})^{3/4}}.$$ Inserting it into the expression for $$I$$ we have

$$I=-\int\frac{t^{3}}{(1+t^{4})^{3/4}}\,dt.$$

Now we observe that the numerator $$t^{3}$$ is (up to a constant) the derivative of the inside of the bracket in the denominator, suggesting the elementary substitution $$z=1+t^{4}.$$ Before making that substitution formally, let us recall the differentiation formula we will need:

For any differentiable function $$z(t)$$,

$$\frac{d}{dt}\Bigl(z(t)\Bigr)^{1/4} =\frac14\,z(t)^{-3/4}\,z'(t).$$

Taking $$z(t)=1+t^{4}$$ gives $$z'(t)=4t^{3},$$ so

$$\frac{d}{dt}\bigl(1+t^{4}\bigr)^{1/4} =\frac14\,(1+t^{4})^{-3/4}\cdot4t^{3} =t^{3}(1+t^{4})^{-3/4}.$$

This derivative is exactly the integrand (up to the overall minus-sign we already have). Therefore

$$-\int t^{3}(1+t^{4})^{-3/4}\,dt =-\Bigl(1+t^{4}\Bigr)^{1/4}+C.$$

We now back-substitute $$t=\frac1x$$ to return to the original variable:

$$1+t^{4}=1+\frac{1}{x^{4}} =\frac{x^{4}+1}{x^{4}},$$

so

$$\Bigl(1+t^{4}\Bigr)^{1/4} =\left(\frac{x^{4}+1}{x^{4}}\right)^{1/4}.$$

Consequently

$$I=-\left(\frac{x^{4}+1}{x^{4}}\right)^{1/4}+C.$$

The antiderivative obtained matches exactly the expression given in Option A.

Hence, the correct answer is Option A.