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# Find the value of the integral of the function$\dfrac{{{\left( 2+{{x}^{2}} \right)}^{2}}{{x}^{3}}}{\left( 1+{{x}^{2}} \right)}$

Last updated date: 21st Jun 2024
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Hint: We solve this question by first considering the given integral. Then we assume that $1+{{x}^{2}}=t$. Then we differentiate it and find the value of $dx$ in terms of $dt$. Then we convert the integral in terms of x into the integral with $t$ as variable and simplify it using the formula, ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$. Then we consider the obtained integral and then solve it using the formula for integration, $\int{{{x}^{n}}dx=\dfrac{{{x}^{n+1}}}{n+1}+c}$ and another formula for integration, $\int{\dfrac{1}{x}dx}=\log x+c$. Then we get the value of integral in terms of variable $t$. Then we substitute the value of $t$ we have assumed in the start, that is $t=1+{{x}^{2}}$ and substitute it in the obtained value to get the final answer.

Now let us consider given the integral
$\int{\dfrac{{{\left( 2+{{x}^{2}} \right)}^{2}}{{x}^{3}}}{\left( 1+{{x}^{2}} \right)}dx}$
Now, let us assume that $1+{{x}^{2}}=t$.
Now let us differentiate it. Then we get,
\begin{align} & \Rightarrow 2xdx=dt \\ & \Rightarrow xdx=\dfrac{dt}{2} \\ \end{align}
Now, we can write the integral as,
\begin{align} & \Rightarrow \int{\dfrac{{{\left( 2+{{x}^{2}} \right)}^{2}}{{x}^{3}}}{\left( 1+{{x}^{2}} \right)}dx}=\int{\dfrac{{{\left( 2+{{x}^{2}} \right)}^{2}}{{x}^{2}}}{\left( 1+{{x}^{2}} \right)}xdx} \\ & \Rightarrow \int{\dfrac{{{\left( 2+{{x}^{2}} \right)}^{2}}{{x}^{3}}}{\left( 1+{{x}^{2}} \right)}dx}=\int{\dfrac{{{\left( 1+t \right)}^{2}}\left( t-1 \right)}{t}\dfrac{dt}{2}} \\ \end{align}
So, we get the integral as,
$\Rightarrow \int{\dfrac{{{\left( 2+{{x}^{2}} \right)}^{2}}{{x}^{3}}}{\left( 1+{{x}^{2}} \right)}dx}=\int{\dfrac{{{\left( 1+t \right)}^{2}}\left( t-1 \right)}{2t}dt}...........\left( 1 \right)$
Now, let us consider the integral $\int{\dfrac{{{\left( 1+t \right)}^{2}}\left( t-1 \right)}{2t}dt}$.
Now let us consider the formula,
${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$
Using this we can simplify the above integral as,
\begin{align} & \Rightarrow \int{\dfrac{{{\left( 1+t \right)}^{2}}\left( t-1 \right)}{2t}dt}=\int{\dfrac{\left( 1+{{t}^{2}}+2t \right)\left( t-1 \right)}{2t}dt} \\ & \Rightarrow \int{\dfrac{{{\left( 1+t \right)}^{2}}\left( t-1 \right)}{2t}dt}=\int{\dfrac{t+{{t}^{3}}+2{{t}^{2}}-1-{{t}^{2}}-2t}{2t}dt} \\ & \Rightarrow \int{\dfrac{{{\left( 1+t \right)}^{2}}\left( t-1 \right)}{2t}dt}=\int{\dfrac{{{t}^{3}}+{{t}^{2}}-t-1}{2t}dt} \\ \end{align}
Simplifying it we get,
\begin{align} & \Rightarrow \int{\dfrac{{{\left( 1+t \right)}^{2}}\left( t-1 \right)}{2t}dt}=\dfrac{1}{2}\int{\dfrac{{{t}^{3}}+{{t}^{2}}-t-1}{t}dt} \\ & \Rightarrow \int{\dfrac{{{\left( 1+t \right)}^{2}}\left( t-1 \right)}{2t}dt}=\dfrac{1}{2}\int{\left( {{t}^{2}}+t-1-\dfrac{1}{t} \right)dt} \\ \end{align}
We can separate the polynomial on the right-hand side of integral and write it as,
$\Rightarrow \int{\dfrac{{{\left( 1+t \right)}^{2}}\left( t-1 \right)}{2t}dt}=\dfrac{1}{2}\left( \int{{{t}^{2}}dt}+\int{tdt}-\int{1dt}-\int{\dfrac{1}{t}dt} \right)$
Now, let us consider the formulas for integration,
\begin{align} & \int{{{x}^{n}}dx=\dfrac{{{x}^{n+1}}}{n+1}+c} \\ & \int{\dfrac{1}{x}dx}=\log x+c \\ \end{align}
Using it we can write the above equation as,
\begin{align} & \Rightarrow \int{\dfrac{{{\left( 1+t \right)}^{2}}\left( t-1 \right)}{2t}dt}=\dfrac{1}{2}\left( \dfrac{{{t}^{3}}}{3}+\dfrac{{{t}^{2}}}{2}-t-\log t \right)+c \\ & \Rightarrow \int{\dfrac{{{\left( 1+t \right)}^{2}}\left( t-1 \right)}{2t}dt}=\dfrac{{{t}^{3}}}{6}+\dfrac{{{t}^{2}}}{4}-\dfrac{t}{2}-\dfrac{\log t}{2}+c \\ \end{align}
Now let us substitute this value in equation (1). Then we get,
$\Rightarrow \int{\dfrac{{{\left( 2+{{x}^{2}} \right)}^{2}}{{x}^{3}}}{\left( 1+{{x}^{2}} \right)}dx}=\dfrac{{{t}^{3}}}{6}+\dfrac{{{t}^{2}}}{4}-\dfrac{t}{2}-\dfrac{\log t}{2}+c$
Now, let us substitute the value of t in terms of x, that is $1+{{x}^{2}}=t$.
Then we get the value of the given integral as,
$\Rightarrow \int{\dfrac{{{\left( 2+{{x}^{2}} \right)}^{2}}{{x}^{3}}}{\left( 1+{{x}^{2}} \right)}dx}=\dfrac{{{\left( 1+{{x}^{2}} \right)}^{3}}}{6}+\dfrac{{{\left( 1+{{x}^{2}} \right)}^{2}}}{4}-\dfrac{\left( 1+{{x}^{2}} \right)}{2}-\dfrac{\log \left( 1+{{x}^{2}} \right)}{2}+c$

Hence the answer is $\dfrac{{{\left( 1+{{x}^{2}} \right)}^{3}}}{6}+\dfrac{{{\left( 1+{{x}^{2}} \right)}^{2}}}{4}-\dfrac{\left( 1+{{x}^{2}} \right)}{2}-\dfrac{\log \left( 1+{{x}^{2}} \right)}{2}+c$.

Note: The common mistake that one makes while solving this problem is after assuming that $1+{{x}^{2}}=t$, one might forget to differentiate it and convert $dx$ to $dt$ and write the integral as,
$\Rightarrow \int{\dfrac{{{\left( 2+{{x}^{2}} \right)}^{2}}{{x}^{3}}}{\left( 1+{{x}^{2}} \right)}dx}=\int{\dfrac{{{\left( 1+t \right)}^{2}}{{\left( t-1 \right)}^{\dfrac{3}{2}}}}{t}dt}$
So, one must remember to find the value of $dx$ and convert it to $dt$ and then solve the integral.