Question

Value of $\int \frac{dx}{x^2{(x^4+1)}^{3/4}}$ is :

• A. $-{(1+\frac{1}{x^4})}^{\frac{1}{4}}+c$
• B. ${(1+\frac{1}{x^4})}^{\frac{1}{4}}+c$
• C. $-{(1-\frac{1}{x^4})}^{\frac{1}{4}}+c$
• D. $Noneofthese$

Answer 1

Answer: (A)

$-{(1+\frac{1}{x^4})}^{\frac{1}{4}}+c$

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

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

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

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

$=\int {\frac{dx}{x^5(\frac{1}{x^4}+1)}}^{\frac{3}{4}}$

Let $t=\frac{1}{x^4}+1\Rightarrow dt=\frac{-4}{x^5}dx\Rightarrow \frac{-dt}{4}=\frac{dx}{x^5}$

$I=\int \frac{-dt}{4t^{\frac{3}{4}}}$

$=\frac{-1}{4}\int \frac{-dt}{t^{\frac{3}{4}}}$

$=\frac{-1}{4}\int t^{\frac{-3}{4}}dt$

$=\frac{-1}{4}\left[\frac{t^{\frac{-3}{4}+1}}{\frac{-3}{4}+1}\right]$

$=\frac{-1}{4}\frac{t^{\frac{1}{4}}}{\frac{1}{4}}$

$=-t^{\frac{1}{4}}+c$ where $c$ is the constant of integration

$\therefore I=-{(\frac{1}{x^4}+1)}^{\frac{1}{4}}+c$ where $t=\frac{1}{x^4}+1$