Question

What is the integral of $\int{\left( \dfrac{1}{{{x}^{2}}+x+1} \right)}dx$ ?

Answer 1

We must express the denominator in the form of a square and a term in addition by replacing ${{x}^{2}}+x+1$ with ${{\left( x+\dfrac{1}{2} \right)}^{2}}+\dfrac{3}{4}$ . We must then use the substitution process for $x+\dfrac{1}{2}$ , and then use the identity of integration, $\int{\dfrac{1}{{{a}^{2}}+{{x}^{2}}}dx=\dfrac{1}{a}{{\tan }^{-1}}}\dfrac{x}{a}$ to get to the final result.

Complete step-by-step solution:
We know that the identity for square of addition of two quantities is expressed as ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$ .
Using this identity, we can write,
${{\left( x+\dfrac{1}{2} \right)}^{2}}={{x}^{2}}+2\times x\times \dfrac{1}{2}+{{\left( \dfrac{1}{2} \right)}^{2}}$
And thus, we get the following equation,
${{\left( x+\dfrac{1}{2} \right)}^{2}}={{x}^{2}}+x+\dfrac{1}{4}$.
Hence using the above equation, we can write
${{x}^{2}}+x+1={{x}^{2}}+x+\dfrac{1}{4}+\dfrac{3}{4}$
And hence, we get
${{x}^{2}}+x+1={{\left( x+\dfrac{1}{2} \right)}^{2}}+\dfrac{3}{4}$
So, now we can now write the given integral as
$I=\int{\left( \dfrac{1}{{{x}^{2}}+x+1} \right)}dx$
$\Rightarrow I=\int{\left( \dfrac{1}{{{\left( x+\dfrac{1}{2} \right)}^{2}}+\dfrac{3}{4}} \right)}dx$
Let us assume a variable $y=x+\dfrac{1}{2}$ .
Let us differentiate this to find the value of dx.
Hence, we get
$dy=dx+0$
Thus $dy=dx.$
And so, our integral now takes the following form,
$\Rightarrow I=\int{\left( \dfrac{1}{{{y}^{2}}+\dfrac{3}{4}} \right)}dy$
We can also write this equation as
$I=\int{\left( \dfrac{1}{{{\left( \dfrac{\sqrt{3}}{2} \right)}^{2}}+{{y}^{2}}} \right)}dy$
We all know very well the following identity for integration,
$\int{\dfrac{1}{{{a}^{2}}+{{x}^{2}}}dx=\dfrac{1}{a}{{\tan }^{-1}}}\dfrac{x}{a}$
Using this identity, we can integrate our integral very easily with $a=\dfrac{\sqrt{3}}{2}$ and replacing x with y. Thus, we get the following equation
$I=\dfrac{1}{\left( \dfrac{\sqrt{3}}{2} \right)}{{\tan }^{-1}}\left( \dfrac{y}{\dfrac{\sqrt{3}}{2}} \right)$
Now, we can substitute the value of y to get the following equation,
$I=\dfrac{1}{\left( \dfrac{\sqrt{3}}{2} \right)}{{\tan }^{-1}}\left( \dfrac{x+\dfrac{1}{2}}{\dfrac{\sqrt{3}}{2}} \right)$
We can again simplify the above equation to get
$I=\dfrac{2}{\sqrt{3}}{{\tan }^{-1}}\left( \dfrac{2x+1}{\sqrt{3}} \right)$
Thus, the integration of $\int{\left( \dfrac{1}{{{x}^{2}}+x+1} \right)}dx$ is $\dfrac{2}{\sqrt{3}}{{\tan }^{-1}}\left( \dfrac{2x+1}{\sqrt{3}} \right)$.

Note: The students must analyse and then decide the approach to solve such problems. We must not try to solve this problem using the partial fraction approach as it will get too complex to solve. We must clearly remember all the formulae for integration as that is the point where most of the students make mistakes.