Question

Find the value of integral $\int{\dfrac{dx}{1+x+{{x}^{2}}}}$.

Answer 1

Hint: First, simplify the denominator using completing the square method to get $\int{\dfrac{dx}{{{\left( x+\dfrac{1}{2} \right)}^{2}}+\dfrac{3}{4}}}$. Next use integration by substitution and let $\dfrac{3}{4}{{u}^{2}}={{\left( x+\dfrac{1}{2} \right)}^{2}}$. Using this we get $\dfrac{2\sqrt{3}}{3}\int{\dfrac{du}{{{u}^{2}}+1}}$. Solve this and then substitute u using $\dfrac{3}{4}{{u}^{2}}={{\left( x+\dfrac{1}{2} \right)}^{2}}$.

Complete step-by-step answer:
In this question, we need to find the value of the integral $\int{\dfrac{dx}{1+x+{{x}^{2}}}}$.
 For solving this question, let us first simplify the given expression.
We will simplify the denominator $1+x+{{x}^{2}}$ using completing the square method.
Let us first see what completing the square method actually is:
In elementary algebra, completing the square method is a technique for converting a quadratic polynomial which is of the form $a{{x}^{2}}+bx+c$ to the form $a{{\left( x-h \right)}^{2}}+k$ for some values of h and k.
So, using this completing the square method, we can write $1+x+{{x}^{2}}$ as:
$1+x+{{x}^{2}}={{\left( x+\dfrac{1}{2} \right)}^{2}}+\dfrac{3}{4}$
Substituting the above expression in the given integral, we will get the following:
$\int{\dfrac{dx}{{{\left( x+\dfrac{1}{2} \right)}^{2}}+\dfrac{3}{4}}}$
Now, we will use integration by substitution.
In calculus, integration by substitution, also known as u-substitution or change of variables, is a method for evaluating integrals. Direct application of the fundamental theorem of calculus to find an antiderivative can be quite difficult, and integration by substitution can help simplify that task. It is the counterpart to the chain rule for differentiation, in fact, it can loosely be thought of as using the chain rule "backwards."
So, we will use integration by substitution in the following way:
Let $\dfrac{3}{4}{{u}^{2}}={{\left( x+\dfrac{1}{2} \right)}^{2}}$
$\dfrac{\sqrt{3}}{2}u=x+\dfrac{1}{2}$
On differentiating this, we will get the following:
$\dfrac{\sqrt{3}}{2}du=dx$
Using this in the above integral, we will get the following:
$\dfrac{\sqrt{3}}{2}\int{\dfrac{du}{\tfrac{3}{4}{{u}^{2}}+\tfrac{3}{4}}}$
$\dfrac{2\sqrt{3}}{3}\int{\dfrac{du}{{{u}^{2}}+1}}$
Now, we know that $\int{\dfrac{du}{{{u}^{2}}+1}}={{\tan }^{-1}}u+c$
Using the above fact, we will get the following:
$\dfrac{2\sqrt{3}}{3}\int{\dfrac{du}{{{u}^{2}}+1}}=\dfrac{2\sqrt{3}}{3}{{\tan }^{-1}}u+c$
$\dfrac{2\sqrt{3}}{3}\int{\dfrac{du}{{{u}^{2}}+1}}=\dfrac{2\sqrt{3}}{3}{{\tan }^{-1}}\left( \dfrac{2x+1}{\sqrt{3}} \right)+c$
Hence, $\int{\dfrac{dx}{1+x+{{x}^{2}}}}=\dfrac{2\sqrt{3}}{3}{{\tan }^{-1}}\left( \dfrac{2x+1}{\sqrt{3}} \right)+c$
This is our final answer.

Note: In this question, it is very important to know about the completing the square method and the integration by substitution method. Instead of integration by substitution, we can also use the direct formula $\int{\dfrac{dx}{{{\left( x+a \right)}^{2}}+{{b}^{2}}}}=\dfrac{1}{b}{{\tan }^{-1}}\left( \dfrac{x+a}{b} \right)+c$ to arrive at the final answer.