Question

What is $\int {\dfrac{{{x^2} - 1}}{{{x^4} + {x^2} + 1}}} dx$ ?

Answer 1

We will assume $I = \int {\dfrac{{{x^2} - 1}}{{{x^4} + {x^2} + 1}}} dx$. Then we will take ${x^2}$ common from the numerator and the denominator. We will make a perfect square of ${\left( {x + \dfrac{1}{x}} \right)^2}$by adding and subtracting $1$ in the denominator. Then we will substitute $x + \dfrac{1}{x} = t$ and then we will integrate it and substitute back $x + \dfrac{1}{x} = t$. We will simplify this to find the result.

Complete step by step answer:
Let $I = \int {\dfrac{{{x^2} - 1}}{{{x^4} + {x^2} + 1}}} dx$. Taking ${x^2}$ common from the numerator and the denominator, we get
$\Rightarrow I = \int {\dfrac{{{x^2}\left( {1 - \dfrac{1}{{{x^2}}}} \right)}}{{{x^2}\left( {{x^2} + 1 + \dfrac{1}{{{x^2}}}} \right)}}} dx$
On cancelling the common term from the numerator and the denominator, we get
$\Rightarrow I = \int {\dfrac{{1 - \dfrac{1}{{{x^2}}}}}{{{x^2} + 1 + \dfrac{1}{{{x^2}}}}}} dx$
Now, adding and subtracting $1$ in the denominator, we get
$\Rightarrow I = \int {\dfrac{{1 - \dfrac{1}{{{x^2}}}}}{{{x^2} + 2 + \dfrac{1}{{{x^2}}} - 1}}} dx$
We can see that $\left( {{x^2} + 2 + \dfrac{1}{{{x^2}}}} \right)$ is equal to ${\left( {x + \dfrac{1}{x}} \right)^2}$.

On rewriting, we get
$\Rightarrow I = \int {\dfrac{{\left( {1 - \dfrac{1}{{{x^2}}}} \right)}}{{{{\left( {x + \dfrac{1}{x}} \right)}^2} - 1}}} dx - - - (1)$
Let $x + \dfrac{1}{x} = t$. On differentiating w.r.t $x$, we get
$\Rightarrow 1 - \dfrac{1}{{{x^2}}} = \dfrac{{dt}}{{dx}}$
$\Rightarrow dt = \left( {1 - \dfrac{1}{{{x^2}}}} \right)dx$
On substituting in $(1)$, we get
$\Rightarrow I = \int {\dfrac{{dt}}{{{t^2} - 1}}}$
Using the identity: ${a^2} - {b^2} = (a + b)(a - b)$, we get
$\Rightarrow I = \int {\dfrac{{dt}}{{\left( {t + 1} \right)\left( {t - 1} \right)}}}$

Multiplying $2$ in the numerator and the denominator, we get
$\Rightarrow I = \dfrac{1}{2}\int {\dfrac{2}{{\left( {t + 1} \right)\left( {t - 1} \right)}}} dt$
On rewriting the numerator, we get
$\Rightarrow I = \dfrac{1}{2}\int {\dfrac{{\left( {t + 1} \right) - \left( {t - 1} \right)}}{{\left( {t + 1} \right)\left( {t - 1} \right)}}} dt$
On dividing, we get
$\Rightarrow I = \dfrac{1}{2}\int {\left( {\dfrac{1}{{t - 1}} - \dfrac{1}{{t + 1}}} \right)} dt$
$\Rightarrow I = \dfrac{1}{2}\left( {\int {\dfrac{1}{{t - 1}}dt} - \int {\dfrac{1}{{t + 1}}dt} } \right)$
On integration, we get
$\Rightarrow I = \dfrac{1}{2}\left[ {\ln \left( {t - 1} \right) - \ln \left( {t + 1} \right)} \right] + C$

From the property of logarithm, we know that $\ln a - \ln b = \ln \left( {\dfrac{a}{b}} \right)$. Using this, we get
$\Rightarrow I = \dfrac{1}{2}\ln \left( {\dfrac{{t - 1}}{{t + 1}}} \right) + C$
Substituting back $x + \dfrac{1}{x} = t$, we get
$\Rightarrow I = \dfrac{1}{2}\ln \left( {\dfrac{{x + \dfrac{1}{x} - 1}}{{x + \dfrac{1}{x} + 1}}} \right) + C$
On simplifying, we get
$\Rightarrow I = \dfrac{1}{2}\ln \left( {\dfrac{{\dfrac{{{x^2} - x + 1}}{x}}}{{\dfrac{{{x^2} + x + 1}}{x}}}} \right) + C$
On further simplification, we get
$\therefore I = \dfrac{1}{2}\ln \left( {\dfrac{{{x^2} - x + 1}}{{{x^2} + x + 1}}} \right) + C$

Therefore, $\int {\dfrac{{{x^2} - 1}}{{{x^4} + {x^2} + 1}}} dx = \dfrac{1}{2}\ln \left( {\dfrac{{{x^2} - x + 1}}{{{x^2} - x + 1}}} \right) + C$.

Note: Here, $C$ is the constant of integration. In a definite integral we get the value of integration constant by using the upper and lower limit of integration. Note, we can also use inverse trigonometric properties to find the antiderivative. Using the direct integration formula that is mentioned as identity can make the problem simple by just substituting those values.