How to evaluate $\int \frac{1}{x^{2}\sqrt[4]{(x^4+1)^3}} dx$

Question

\begin{align*} &\int \frac{1}{x^{2}\sqrt[4]{(x^4+1)^3}} dx \\ \end{align*}

\begin{aligned} &\int \frac{1}{x^{2}\sqrt[4]{(x^4+1)^3}} dx \\ &= \int \frac{1}{x^{2}\sqrt[4]{u^3}} \cdot \frac{1}{4x^3} du \\ &= \frac{1}{4} \int \frac{1}{u^{3/4}} du \\ &= -\frac{1}{12} u^{-1/4} + C \\ &= -\frac{1}{12} (1 + x^4)^{-1/4} + C \end{aligned}

Answer 1

Notice $$\begin{align} \frac{1}{x^2 (x^4+1)^{3/4}} &= \frac{x^4 - x^4 + 1}{x^2 (x^4+1)^{3/4}} \\ &= -\frac{x^4}{x^2(x^4+1)^{3/4}} + \frac{x^4+1}{x^2(x^4+1)^{3/4}} \\ &= -\frac{x^3}{x(x^4+1)^{3/4}} + \frac{(x^4+1)^{1/4}}{x^2} \tag{1}. \end{align}$$ Now recalling the quotient rule $$\frac{d}{dx}\left[\frac{f(x)}{g(x)}\right] = \frac{f'(x)}{g(x)} - \frac{f(x)g'(x)}{g(x)^2},$$ with the choice $$f(x) = (x^4+1)^{1/4}, \quad f'(x) = \frac{x^3}{(x^4+1)^{3/4}}, \\ g(x) = -x, \quad g'(x) = -1,$$ we see that $$\frac{d}{dx}\left[-\frac{(x^4+1)^{1/4}}{x}\right] = -\frac{x^3}{x(x^4+1)^{3/4}} + \frac{(x^4+1)^{1/4} (-1)}{(-x)^2}.$$ Therefore, $$\int \frac{dx}{x^2(x^4+1)^{3/4}} = \int \frac{d}{dx}\left[-\frac{(x^4+1)^{1/4}}{x}\right] \, dx = -\frac{(x^4+1)^{1/4}}{x} + C.$$

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If you don't like the quotient rule trick, we can proceed with integration by parts of $(1)$:

$$ I = \int \frac{dx}{x^2(x^4+1)^{3/4}} = -\int \frac{x^3}{x(x^4+1)^{3/4}} \, dx + \int \frac{(x^4+1)^{1/4}}{x^2} \, dx. \tag{2}$$ Let $$J = \int \frac{(x^4+1)^{1/4}}{x^2} \, dx.$$ Then with the choice $$u = \frac{1}{x}, \quad du = -\frac{1}{x^2} \, dx, \\ dv = \frac{x^3}{(x^4+1)^{3/4}} \, dx, \quad v = (x^4+1)^{1/4},$$ we obtain $$I = -\left(\frac{(x^4+1)^{1/4}}{x} + \int \frac{(x^4+1)^{1/4}}{x^2} \, dx \right) + J = -\frac{(x^4+1)^{1/4}}{x} - J + J + C$$ which is the same as above.

The key in any case is to recognize how to rewrite the integrand in the form shown in Equation $(1)$.

Answer 2

$$ \begin{aligned} \int \frac{1}{x^2\left(x^4+1\right)^{\frac{2}{4}}} d x = & \int \frac{\frac{1}{x^5}}{\left(1+\frac{1}{x^4}\right)^{\frac{3}{4}}} d x \\ = & -\frac{1}{4} \int \frac{d\left(1+\frac{1}{x^4}\right)}{\left(1+\frac{1}{4}\right)^{\frac{3}{4}}}\\=&-\left(1+\frac{1}{x^4}\right)^{\frac{1}{4}}+C\\OR=& -\frac{\left(x^4+1\right)^{\frac{1}{4}}}{x}+C \end{aligned} $$

Answer 3

Hint The appearance of the factor $(x^4 + 1)^a$ suggests the substitution $$x^2 = \tan \theta, \qquad 2 x \,dx = \sec^2 \theta \,d\theta,$$ so that factor is $\sec^{2 a} \theta$, giving the integral $$\frac12 \int \frac{\cos \theta \,d\theta}{\sin^{3 / 2} \theta} ,$$ which can in turn be handled with a straightforward substitution and the power rule.

Answer 4

$$I:=\int \frac{dx}{x^2 \sqrt[4]{(x^4+1)^3}}$$ let $u=\frac{1}{x} $ then $du = \frac{-dx}{x^2} $ $$I=\int \frac{-du}{\sqrt[4]{(\frac{1}{u^4}+1)^3}}=\int \frac{-du}{\sqrt[4]{(\frac{1+u^4}{u^4})^3}}= \int \frac{-u^3 du}{\sqrt[4]{({1+u^4})^3}}$$ it is obvious that we can get rid of $u^4$ by substituting again let $u^4 =t$ then $4u^3 = dt $ and notice $t={\frac{1}{x^4}}$ $$I=\int \frac{-dt}{4\sqrt[4]{(1+t)^3}} = \frac{-1}{4} *4 (1+t)^{\frac{1}{4}}+C =-(1+\frac{1}{x^4})^{\frac{1}{4}}+C =-\frac{\sqrt[4]{(1+x^4)}}{x}+C$$