Question

Evaluate the following integral:
$\int{\dfrac{{{x}^{2}}+9}{{{x}^{4}}+81}dx}$

Answer 1

Hint: We will be using the concepts of integral calculus to solve the problem. We will first divide the numerator and denominator by x square and then we will convert the denominator by using completing the square then we will integrate the integral obtained by using the substitution method. We will be using some algebraic identities to solve the problem like,
$\begin{align}
  & {{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab \\
 & {{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab \\
 & {{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right) \\
\end{align}$

Complete step-by-step answer:
Now, we have to evaluate,
$\int{\dfrac{{{x}^{2}}+9}{{{x}^{4}}+81}dx}$
Now, we can write it as,
$\int{\dfrac{{{x}^{2}}+9}{{{\left( {{x}^{2}} \right)}^{2}}+{{\left( 9 \right)}^{2}}}dx}$
Now, we will divide the numerator and denominator by ${{x}^{2}}$. So, that we have,
$\int{\dfrac{1+\dfrac{9}{{{x}^{2}}}}{{{x}^{2}}+\dfrac{{{9}^{2}}}{{{x}^{2}}}}dx}$
Now, we know that,
$\dfrac{d}{dx}\left( x-\dfrac{9}{x} \right)=1+\dfrac{9}{{{x}^{2}}}$
Also, $1+\dfrac{9}{{{x}^{2}}}$ is the numerator in integral. So, we let,
$y=x-\dfrac{9}{x}$
Now, we can write denominator in integral as,
$\begin{align}
  & {{\left( x \right)}^{2}}+{{\left( \dfrac{9}{x} \right)}^{2}}={{\left( x \right)}^{2}}+{{\left( \dfrac{9}{x} \right)}^{2}}-2\times \dfrac{9}{x}\times x+18 \\
 & {{x}^{2}}+{{\left( \dfrac{9}{x} \right)}^{2}}={{\left( x-\dfrac{9}{x} \right)}^{2}}+18 \\
\end{align}$
So, now we have the integral as,
$\begin{align}
  & \int{\dfrac{\left( 1+\dfrac{9}{{{x}^{2}}} \right)}{{{\left( x-\dfrac{9}{x} \right)}^{2}}+18}dx} \\
 & \Rightarrow y=x-\dfrac{9}{x} \\
 & \Rightarrow dy=\left( 1+\dfrac{9}{{{x}^{2}}} \right)dx \\
\end{align}$
So, we have the integral on substituting this as,
$\int{\dfrac{dy}{{{y}^{2}}+18}}$
Now, we know a standard integral that $\int{\dfrac{dx}{{{x}^{2}}+{{a}^{2}}}=\dfrac{1}{a}{{\tan }^{-1}}\left( \dfrac{x}{a} \right)+c}$.
So, we will use this to solve,
$\begin{align}
  & \int{\dfrac{dy}{{{y}^{2}}+18}}=\int{\dfrac{dy}{{{y}^{2}}+{{\left( 3\sqrt{2} \right)}^{2}}}} \\
 & =\dfrac{1}{3\sqrt{2}}{{\tan }^{-1}}\left( \dfrac{y}{3\sqrt{2}} \right)+c \\
\end{align}$
Now, we again put the value of $y=x-\dfrac{9}{x}$. So, that we have,
$\int{\dfrac{{{x}^{2}}+9}{{{x}^{4}}+81}dx}=\dfrac{1}{3\sqrt{2}}{{\tan }^{-1}}\left( \dfrac{{{x}^{2}}-9}{3\sqrt{2}x} \right)+c$
Where c is constant of integration.

Note: To solve these type of questions it is important to remember some basic concepts of integral calculus like integral by substitution also it has to be noted that how we have chosen to substitute $y=x-\dfrac{9}{x}$ by converting the denominator as
  $\begin{align}
  & {{\left( x \right)}^{2}}+{{\left( \dfrac{9}{x} \right)}^{2}}={{\left( x \right)}^{2}}+{{\left( \dfrac{9}{x} \right)}^{2}}-2\times \dfrac{9}{x}\times x+18 \\
 & {{x}^{2}}+{{\left( \dfrac{9}{x} \right)}^{2}}={{\left( x-\dfrac{9}{x} \right)}^{2}}+18 \\
\end{align}$
So that the integral got converted into a single variable.