Question

If $\int{\dfrac{dx}{{{x}^{2}}{{\left( 1+{{x}^{4}} \right)}^{\dfrac{3}{4}}}}}=f\left( x \right){{\left( 1+{{x}^{4}} \right)}^{2}}+C$ where, $C$ is a constant of integration, then find the function $f\left( x \right)$. $$$$

Answer 1

We proceed from left hand side and take ${{x}^{4}}$common from the bracket. We substitute $t=\dfrac{1}{{{x}^{4}}}+1$ and find $dx$ in terms of $t,dt$ also substituted in the integrand. We integrate with respect to $t$ and put back $t$ in terms of $x$. We multiply and divide $x$and then compare the resultant expression with expression at the right hand side to get $f\left( x \right)$. $$$$

Complete step by step answer:
We are given in the question an equation whose left hand side has an integral and the right hand side has functional equation as
$\int{\dfrac{dx}{{{x}^{2}}{{\left( 1+{{x}^{4}} \right)}^{\dfrac{3}{4}}}}}=f\left( x \right){{\left( 1+{{x}^{4}} \right)}^{2}}+C$
If we shall integrate the left hand side and try to express the result similar to the expression at the right hand side we may get the required function $f\left( x \right)$. We proceed from left hand side,
$\Rightarrow \int{\dfrac{dx}{{{x}^{2}}{{\left( 1+{{x}^{4}} \right)}^{\dfrac{3}{4}}}}}$
We take ${{x}^{4}}$ common from the bracket in the denominator to get,

& \Rightarrow \int{\dfrac{dx}{{{x}^{2}}{{\left( {{x}^{4}}\left( \dfrac{1}{{{x}^{4}}}+1 \right) \right)}^{\dfrac{3}{4}}}}} \\\ & \Rightarrow \int{\dfrac{dx}{{{x}^{2}}{{\left( {{x}^{4}} \right)}^{\dfrac{3}{4}}}{{\left( \dfrac{1}{{{x}^{4}}}+1 \right)}^{\dfrac{3}{4}}}}} \\\ \end{aligned}$$ We use the exponential identity of power raised to another power ${{\left( {{a}^{m}} \right)}^{n}}={{a}^{m\times n}},a\ne 0$ for $a=x,m=4,n=\dfrac{3}{4}$ in the above step to have $$\begin{aligned} & \Rightarrow \int{\dfrac{dx}{{{x}^{2}}{{x}^{4\times \dfrac{3}{4}}}{{\left( \dfrac{1}{{{x}^{4}}}+1 \right)}^{\dfrac{3}{4}}}}} \\\ & \Rightarrow \int{\dfrac{dx}{{{x}^{2}}{{x}^{3}}{{\left( \dfrac{1}{{{x}^{4}}}+1 \right)}^{\dfrac{3}{4}}}}} \\\ \end{aligned}$$ We use the exponential identity of power product with same base ${{a}^{m}}\times {{a}^{n}}={{a}^{m+n}}$ for $a=x,m=3,n=4$ in the above step to have, $$\begin{aligned} & \Rightarrow \int{\dfrac{dx}{{{x}^{5}}{{\left( \dfrac{1}{{{x}^{4}}}+1 \right)}^{\dfrac{3}{4}}}}} \\\ & \Rightarrow \int{\dfrac{dx}{{{x}^{5}}{{\left( \dfrac{1}{{{x}^{4}}}+1 \right)}^{\dfrac{3}{4}}}}} \\\ & \Rightarrow \int{\dfrac{1}{{{x}^{5}}}}{{\left( \dfrac{1}{{{x}^{4}}}+1 \right)}^{-\dfrac{3}{4}}}dx.........\left( 1 \right) \\\ \end{aligned}$$ We can substitute $t=\dfrac{1}{{{x}^{4}}}+1$ in the above step since integration is independent of change in variable and for that we also need $dx$ and rest of the expression in $x$ in terms of $t$. We differentiate $t=\dfrac{1}{{{x}^{4}}}+1$ both side with respect to $x$ to have, $$\begin{aligned} & \dfrac{d}{dx}t=\dfrac{d}{dx}\left( \dfrac{1}{{{x}^{4}}}+1 \right)=\dfrac{d}{dx}\left( {{x}^{-4}}+1 \right) \\\ & \Rightarrow \dfrac{dt}{dx}=-4{{x}^{-4-1}}+0 \\\ & \Rightarrow dx=\dfrac{dt}{-4{{x}^{-5}}} \\\ \end{aligned}$$ We put $t=\dfrac{1}{{{x}^{4}}}+1$ and $dx=\dfrac{dt}{-4{{x}^{-5}}}$ in (1) to have $$\begin{aligned} & \Rightarrow \int{\dfrac{1}{{{x}^{5}}}\times t\times \dfrac{dt}{-4{{x}^{-5}}}} \\\ & \Rightarrow \dfrac{-1}{4}\int{\dfrac{tdt}{{{x}^{-5}}{{x}^{5}}}} \\\ \end{aligned}$$ We use the exponential identity of power product with same base ${{a}^{m}}\times {{a}^{n}}={{a}^{m+n}}$ for $a=x,m=-5,n=5$ in the above step to have $$\begin{aligned} & \Rightarrow \dfrac{-1}{4}\int{\dfrac{tdt}{{{x}^{-5+5}}}} \\\ & \Rightarrow \dfrac{-1}{4}\int{tdt} \\\ & \Rightarrow \dfrac{-1}{4}\dfrac{{{t}^{2}}}{2}+C \\\ & \Rightarrow \dfrac{-1}{8}{{t}^{2}}+C \\\ \end{aligned}$$ We put $t=\dfrac{1}{{{x}^{4}}}+1$ in the above step to have, $$\begin{aligned} & \Rightarrow \dfrac{-1}{8}{{\left( \dfrac{1}{{{x}^{4}}}+1 \right)}^{2}}+C \\\ & \Rightarrow \dfrac{-1}{8}{{\left( \dfrac{{{x}^{4}}+1}{{{x}^{4}}} \right)}^{2}}+C \\\ & \Rightarrow \dfrac{-1}{8}\dfrac{{{\left( {{x}^{4}}+1 \right)}^{2}}}{{{\left( {{x}^{4}} \right)}^{2}}}+C \\\ \end{aligned}$$ We use exponential identity of power raised to another power ${{\left( {{a}^{m}} \right)}^{n}}={{a}^{m\times n}},a\ne 0$ for $a=x,m=4,n=2$ in the above step to have $$\Rightarrow \dfrac{-1}{8{{x}^{8}}}{{\left( {{x}^{4}}+1 \right)}^{2}}+C=f\left( x \right){{\left( 1+{{x}^{4}} \right)}^{2}}+C$$ We compare both side and have the required function as $f\left( x \right)=\dfrac{-1}{8{{x}^{8}}}$. $$$$ **Note:** We note that the function is not defined for $x=0$ and the question assumes that. The method of integration we used here is called t-substitution or integration by substitution. We have also used the formula standard integral $\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1},n\ne -1$ and the standard differentiation formula $\dfrac{d}{dx}{{x}^{n}}=n{{x}^{n-1}}$ frequently in this problem.