Question

$\int \frac{1}{x^2{\left(x^4+1\right)}^{{\displaystyle \frac{3}{4}}}}dx=$

• A. $-\frac{{\left(x^4+1\right)}^{\frac{3}{4}}}{x}+C$
• B. $-\frac{{\left(1-x^4\right)}^{\frac{1}{4}}}{2x}+C$
• C. $-\frac{{\left(1+x^4\right)}^{\frac{1}{4}}}{x}+C$
• D. $-\frac{{\left(1+x^4\right)}^{\frac{1}{4}}}{x^2}+C$
• E. $-\frac{{\left(1+x^4\right)}^{\frac{1}{2}}}{x}+C$

Answer 1

The correct option is C

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

Explanation for Correct answer:

Finding the value for the given integral:

$$\begin{gathered} \int \frac{1}{x^2{\left(x^4+1\right)}^{{\displaystyle \frac{3}{4}}}}dx=\int \frac{1}{x^2{\left(x^4\right)}^{\frac{3}{4}}{\left(1+\frac{1}{x^4}\right)}^{\frac{3}{4}}}dx \\ =\int \frac{1}{x^2{x}^3{\left(1+\frac{1}{x^4}\right)}^{\frac{3}{4}}}dx \\ =\int \frac{x^{-5}}{{\displaystyle {\left(1+x^{-4}\right)}^{\frac{3}{4}}}}dx \end{gathered}$$

Solving this by using the substitution method

$$\begin{gathered} t^4=\left(1+x^{-4}\right) \\ 4t^{3}dt=-4x^{-5}dx\left[\because \frac{d}{dx}{x}^n=n{x}^{n-1}\right] \\ {t}^{3}dt=-x^{-5}dx \end{gathered}$$

$$\begin{gathered} \int \frac{x^{-5}}{{\displaystyle {\left(1+x^{-4}\right)}^{\frac{3}{4}}}}dx=-\int \frac{t^{3}dt}{{\left(t^4\right)}^{\frac{3}{4}}} \\ =-\int dt \\ =-t+C \\ =-{\left(1+x^{-4}\right)}^{\frac{1}{4}}+C \\ =-{\left(1+\frac{1}{x^4}\right)}^{\frac{1}{4}}+C \\ =-{\left(\frac{x^4+1}{x^4}\right)}^{\frac{1}{4}}+C \\ =-\frac{{\left(x^4+1\right)}^{\frac{1}{4}}}{{\left(x^4\right)}^{\frac{1}{4}}}+C \\ =-\frac{{\left(x^4+1\right)}^{\frac{1}{4}}}{x}+C \end{gathered}$$

Hence, option (C) is the correct answer.