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

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