Question

Integrate the following fraction$\dfrac{1}{{\left( {{x^2} + 9} \right)}}$.

Answer 1

In the above question we are to find integration of the given fraction. Since it is an indefinite integral that has no upper and lower limit, we will try and convert the given fraction into a form such that standard integral properties can be applied.

Complete step by step solution:
Integration is the inverse process of differentiation. Instead of differentiating a function, we are given derivative of a function and asked to find its primitive, that is, the original function. There are a number of standard integrals and integration properties which are applied to find the final solution.
Given that, $\dfrac{1}{{\left( {{x^2} + 9} \right)}}$
The expression $\left( {{x^2} + 9} \right)$ can be rewritten as,
\Rightarrow \left( {{x^2} + 9} \right) = 9\left\\{ {\left( {\dfrac{x}{9}} \right) + 1} \right\\} = 9\left\\{ {{{\left( {\dfrac{x}{3}} \right)}^2} + 1} \right\\}
Thus, we put this value in our given fraction, we get,
\Rightarrow \dfrac{1}{{\left( {{x^2} + 9} \right)}} = \dfrac{1}{{9\left\\{ {{{\left( {\dfrac{x}{3}} \right)}^2} + 1} \right\\}}} - - - - - \left( 1 \right)
To solve the question further, we will use a standard integral
That is,
$\int {\dfrac{1}{{{u^2} + 1}}du} = {\tan ^{ - 1}}\left( u \right) + C - - - - - \left( 2 \right)$
Taking integration of the (1)
We get,
\Rightarrow \dfrac{1}{9}\int {\dfrac{1}{{\left\\{ {{{\left( {\dfrac{x}{3}} \right)}^2} + 1} \right\\}}}} dx - - - - - \left( 3 \right)
Let, $u = \dfrac{x}{3}$implying $du = \dfrac{1}{3}dx$
Therefore, (3) becomes
\Rightarrow \dfrac{1}{3}\int {\dfrac{{\dfrac{1}{3}}}{{\left\\{ {{{\left( {\dfrac{x}{3}} \right)}^2} + 1} \right\\}}}} dx
$\Rightarrow \dfrac{1}{3}\int {\dfrac{1}{{{u^2} + 1}}du}$
Thus, applying standard integral in (2) in the above expression,
We get,

$\Rightarrow \dfrac{1}{3}{\tan ^{ - 1}}\left( {\dfrac{x}{3}} \right) + C$
Which is our required solution.

Note:
we can solve the given question alternatively in the following manner,
Given that,
$\Rightarrow \dfrac{1}{{\left( {{x^2} + 9} \right)}}$
Taking integration on the above fraction,
We get,
$\Rightarrow \int {\dfrac{{dx}}{{{x^2} + 9}}}$
Let, $x = 3\tan \theta$ implying that $dx = 3{\sec ^2}\theta d\theta$
Therefore, we get,
$\Rightarrow \int {\dfrac{{3{{\sec }^2}\theta d\theta }}{{9\tan \theta + 9}}}$
According to trigonometric identity, we know that
${\tan ^2}\theta + 1 = {\sec ^2}\theta$
Substituting the value of ${\sec ^2}\theta$in the expression, we get
$\Rightarrow \dfrac{1}{3}\int {\dfrac{{{{\sec }^2}\theta }}{{{{\sec }^2}\theta }}d\theta }$
Therefore,
$\Rightarrow \dfrac{1}{3}\int {d\theta }$
$\Rightarrow \dfrac{1}{3}\theta + C$
Reversing the original substitution,
$\Rightarrow x = 3\tan \theta$
Thus we get,
$\Rightarrow \dfrac{1}{3}{\tan ^{ - 1}}\left( {\dfrac{x}{3}} \right) + C$
Which is our required integration.
There can be more than one way of solving any integration, as you can see in the above question. One might have to use more than one method to solve a single question.