Question

Evaluate\[\int {\dfrac{{dx}}{{{x^2} - 6x + 13}}} \].

Answer 1

Hint: For solving these types of integration questions, firstly convert the ${x^2} - 6x + 13$ into complete square and then use appropriate formula, i.e., \[\int {\dfrac{{dx}}{{{x^2} + {a^2}}}} = \dfrac{1}{a}{\tan ^{ - 1}}\left( {\dfrac{x}{a}} \right) + C\] to find out the value of integration.

Complete step-by-step answer:
\[\int {\dfrac{{dx}}{{{x^2} - 6x + 13}}} \]
=\[\int {\dfrac{{dx}}{{{x^2} - 2 \times 3 \times x + 13}}} \]
Add and subtract \[{\left( 3 \right)^2}\] to make complete square,
=\[\int {\dfrac{{dx}}{{\left( {{x^2} - 2 \times 3 \times x + {3^2}} \right) + 13 - {3^2}}}} \]
=\[\int {\dfrac{{dx}}{{{{\left( {x - 3} \right)}^2} + 13 - 9}}} \]
=\[\int {\dfrac{{dx}}{{{{\left( {x - 3} \right)}^2} + 4}}} \]
=\[\int {\dfrac{{dx}}{{{{\left( {x - 3} \right)}^2} + {2^2}}}} \]
It is of the form
\[\int {\dfrac{{dx}}{{{x^2} + {a^2}}}} = \dfrac{1}{a}{\tan ^{ - 1}}\left( {\dfrac{x}{a}} \right) + C\] , where $C$ is the constant of integration.
Replacing $x$ with $\left( {x - 3} \right)$ and $a$ with $2$,
\[\int {\dfrac{{dx}}{{{{\left( {x - 3} \right)}^2} + {2^2}}}} = \dfrac{1}{2}{\tan ^{ - 1}}\left( {\dfrac{{x - 3}}{2}} \right) + C\]

Note: For making the complete square of an equation $a{x^2} \pm bx + c$, we generally add and subtract the term ${\left( {\dfrac{b}{{2a}}} \right)^2}$. Hence for making the complete square of ${x^2} - 6x + 13$, we add and subtract the term ${\left( {\dfrac{{ - 6}}{{2 \times 1}}} \right)^2} = {\left( { - 3} \right)^2} = 9$.