Question

Evaluate \[\int {\dfrac{{x + 2}}{{\sqrt {{x^2} + 5x + 6} }}dx} .\]

Answer 1

Hint: To evaluate the given integral, first of all consider the given integral to be some variable, and then multiply and divide it with $2$ and then with help of algebraic operation, separate the integrand in two terms, then integrate them separately to evaluate the given integral.

Complete step by step solution:
In order to find the integral of the given integrand, let us consider the given integrand to be $I$, that is
$I = \int {\dfrac{{x + 2}}{{\sqrt {{x^2} + 5x + 6} }}dx} $
Now multiplying and dividing the integrand that is left hand side of the equation with $2$ to make the numerator such that it will be equals to the differentiation of denominator, to do this we have to do some algebraic operations too, that will be done further, after this step
$\Rightarrow I = 2 \times \dfrac{1}{2} \times \int {\dfrac{{x + 2}}{{\sqrt {{x^2} + 5x + 6} }}dx} $
We can write it as
$\Rightarrow I = \dfrac{1}{2} \times \int {\dfrac{{2x + 4}}{{\sqrt {{x^2} + 5x + 6} }}dx} $
Now, adding and subtracting $1$ in the numerator of the integrand, we will get
$ I = \dfrac{1}{2} \times \int {\dfrac{{2x + 4 + 1 - 1}}{{\sqrt {{x^2} + 5x + 6} }}dx} $
$ \Rightarrow I = \dfrac{1}{2} \times \int {\dfrac{{2x + 5 - 1}}{{\sqrt {{x^2} + 5x + 6} }}dx} $
We can also write it as
$ I = \dfrac{1}{2} \times \int {\dfrac{{(2x + 5) - 1}}{{\sqrt {{x^2} + 5x + 6} }}dx}$
$\Rightarrow I = \dfrac{1}{2} \times \int {\left( {\dfrac{{2x + 5}}{{\sqrt {{x^2} + 5x + 6} }} - \dfrac{1}{{\sqrt {{x^2} + 5x + 6} }}} \right)dx} $
Using distributive property of integration, we will get
\[ I = \dfrac{1}{2} \times \left( {\int {\dfrac{{2x + 5}}{{\sqrt {{x^2} + 5x + 6} }}dx} - \int {\dfrac{1}{{\sqrt {{x^2} + 5x + 6} }}dx} } \right)\]
Now, let us take ${x^2} + 5x + 6 = t \Rightarrow (2x + 5)dx = dt$ and completing the square of the denominator of the second term integrand,
We can write ${x^2} + 5x + 6 = {x^2} + 2 \times x \times \dfrac{5}{2} + {\left( {\dfrac{5}{2}} \right)^2} - {\left( {\dfrac{5}{2}} \right)^2} + 6 = {\left( {x + \dfrac{5}{2}} \right)^2} - \dfrac{{25}}{4} + 6 = \dfrac{{{{\left( {2x + 5} \right)}^2} - 1}}{4}$
Using above substitutions, we will get
\[ I = \dfrac{1}{2} \times \left( {\int {\dfrac{1}{{\sqrt t }}dt} - \int {\dfrac{1}{{\sqrt {\dfrac{{{{\left( {2x + 5} \right)}^2} - 1}}{4}} }}dx} } \right) \]
$ \Rightarrow I = \dfrac{1}{2} \times \left( {\dfrac{{{t^{ - \dfrac{1}{2} + 1}}}}{{ - \dfrac{1}{2} + 1}} - \int {\dfrac{1}{{\dfrac{{\sqrt {{{\left( {2x + 5} \right)}^2} - 1} }}{2}}}dx} } \right) $
$\Rightarrow I = \dfrac{1}{2} \times \left( {2{t^{\dfrac{1}{2}}} - \int {\dfrac{2}{{\sqrt {{{\left( {2x + 5} \right)}^2} - 1} }}dx} } \right) $
Taking $2x + 5 = u \Rightarrow 2dx = du$
 \[\Rightarrow I = \dfrac{1}{2} \times \left( {2{t^{\dfrac{1}{2}}} + a - \int {\dfrac{1}{{\sqrt {{u^2} - 1} }}du} } \right) \]
$\Rightarrow I = \dfrac{1}{2} \times \left( {2{t^{\dfrac{1}{2}}} + a - \ln \left| {\sqrt {{u^2} - 1} - u} \right| + b} \right)$
$\Rightarrow I = \dfrac{1}{2} \times \left( {2{t^{\dfrac{1}{2}}} - \ln \left| {\sqrt {{u^2} - 1} - u} \right| + a + b} \right) $
Putting the values back, and simplifying further
\[I = {\left( {{x^2} + 5x + 6} \right)^{\dfrac{1}{2}}} - \dfrac{1}{2} \times \ln \left| {\sqrt {{{(2x + 5)}^2} - 1} - (2x + 5)} \right| + \dfrac{1}{2} \times \left( {a + b} \right)\]
Writing \[\dfrac{1}{2} \times \left( {a + b} \right) = c\] and simplifying further, we will get
\[I = {\left( {{x^2} + 5x + 6} \right)^{\dfrac{1}{2}}} - \dfrac{1}{2} \times \ln \left| {\sqrt {{{(2x + 5)}^2} - 1} - (2x + 5)} \right| + c\]
Therefore this is the required integral of the given integrand.
So, the correct answer is “ \[I = {\left( {{x^2} + 5x + 6} \right)^{\dfrac{1}{2}}} - \dfrac{1}{2} \times \ln \left| {\sqrt {{{(2x + 5)}^2} - 1} - (2x + 5)} \right| + c\] ”.

Note: We have written \[\dfrac{1}{2} \times \left( {a + b} \right) = c\] because in indefinite integration, we always get arbitrary constant which means the value will be constant but it will not be fixed, so we can take any variable to present it. Also to make the process easy you can separately integrate both the terms and then subtract them back finally to get the required integration.