Question

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

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

Answer 1

The correct option is A

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

Explanation for Correct answer:

Finding the value of the given integral:

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

Solving this by using the substitution method

$$\begin{gathered} t=\frac{1}{x^2}-x^2 \\ dt=\left(\frac{-2}{x^3}-2x\right)dx\left[\because \frac{d}{dx}{x}^n=n{x}^{n-1}\right] \\ dt=-2\left(\frac{1}{x^3}+x\right)dx \end{gathered}$$

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

Hence, option (A) is the correct answer.