Mathematics

# Integrate $\displaystyle \int{\dfrac{dx}{{x}^{2}{\left(1+{x}^{4}\right)}^{\frac{3}{4}}}}$.

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

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

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

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

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

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

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

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

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

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

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

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