Question

Integrate $\int {\dfrac{x}{{{x^3} + {x^2} + x + 1}}} dx$

Answer 1

Hint: Here we have to integrate the given function so we will use the concept of integration using partial fractions. Firstly, we will factorize the denominator parts into linear factors and then we will find the coefficients and use integration formula such as $\int {\dfrac{1}{x}} dx = \log x$.

Complete answer:
Here, we have the given function is an improper rational fraction then it can be expressed as partial fraction. If $\dfrac{{P(x)}}{{Q(x)}}$ is an improper rational function then it can be expressed as partial fraction as $A(x) + \dfrac{{R(x)}}{{Q(x)}}$ where $A(x)$ is the polynomial in $x$ and $\dfrac{{R(x)}}{{Q(x)}}$ is a proper rational function.
Here, we have to integrate $\int {\dfrac{x}{{{x^3} + {x^2} + x + 1}}} dx$
Now, we will factorize the denominator part of the given fractional function
$ \Rightarrow \int {\dfrac{x}{{{x^3} + {x^2} + x + 1}}} dx = \int {\dfrac{x}{{x({x^2} + x + 1) + 1}}} dx$
The above expression can be written as
$ \Rightarrow \int {\dfrac{x}{{x({x^2} + x + 1) + 1}}} dx = \int {\dfrac{A}{{x + 1}}} + \dfrac{{Bx + C}}{{{x^2} + 1}}$
$ \Rightarrow x = A({x^2} + 1) + (Bx + C)(x + 1)$
On comparing the coefficients of ${x^2}$. We get,
$ \Rightarrow A + B = 0$
$ \Rightarrow A = - B \ldots (1)$
On comparing the coefficients of $x$. We get,
$ \Rightarrow B + C = 1 \ldots (2)$
On comparing the constants parts of both side of the expression. We get,
$ \Rightarrow A + C = 0 \ldots (3)$
Put the value of $A$ from equation $(1)$ in equation $(3)$. We get,
$ \Rightarrow - B + C = 0 \ldots (4)$
Now adding equation $(2)$ and $(4)$ . we get,
$ \Rightarrow 2C = 1$
$ \Rightarrow C = \dfrac{1}{2}$
Putting the value of $C = \dfrac{1}{2}$ in equation $(4)$. We get,
$ \Rightarrow - B + \dfrac{1}{2} = 0$
$ \Rightarrow B = \dfrac{1}{2}$
Putting the value of $B = \dfrac{1}{2}$ in equation $(1)$. We get
$ \Rightarrow A = - \dfrac{1}{2}$
Now putting the value of constants in the function. we get,
$ \Rightarrow \int {\dfrac{x}{{{x^3} + {x^2} + x + 1}}} dx = \int {\left[ { - \dfrac{1}{{2(x + 1)}} + \dfrac{x}{{2({x^2} + 1)}} + \dfrac{1}{{2({x^2} + 1)}}} \right]} dx$
Now we will integrate all the three functions by using the various integration rule and then add their results.
Now, $\int {\dfrac{1}{{2(x + 1)}}dx} = \dfrac{1}{2}\ln (x + 1)$
As we know $\int {\dfrac{1}{x}} dx = \log x$
Now, $\int {\dfrac{x}{{2({x^2} + 1)}}} dx$
Using the formula $\int {\dfrac{{f'(x)}}{{f(x)}}} = \ln \left| {f(x)} \right|$
Now $\dfrac{{d({x^2} + 1)}}{{dx}} = 2x$
Therefore, $\int {\dfrac{x}{{2({x^2} + 1)}}dx} = \dfrac{1}{2}\left( {\dfrac{1}{2}\ln ({x^2} + 1)} \right)$
Now $\int {\dfrac{1}{{2({x^2} + 1)}}dx} = \dfrac{1}{2}{\tan ^{ - 1}}x$
As we know that $\int {\dfrac{1}{{({x^2} + 1)}}} dx = {\tan ^{ - 1}}x$
Now adding the results of all the functions we get,
$ \Rightarrow - \dfrac{1}{2}\ln (x + 1) + \dfrac{1}{2}\left[ {\dfrac{1}{2}\ln ({x^2} + 1)} \right] + \dfrac{1}{2}{\tan ^{ - 1}}x + C$
Hence, the integration of $\int {\dfrac{x}{{{x^3} + {x^2} + x + 1}}} dx$ is $ - \dfrac{1}{2}\ln (x + 1) + \dfrac{1}{2}\left[ {\dfrac{1}{2}\ln ({x^2} + 1)} \right] + \dfrac{1}{2}{\tan ^{ - 1}}x + C$

Note:
Note that we can only use the concept of partial fractions if the degree of the numerator is strictly less than the degree of the denominator. If the degree of numerator is equal or greater than the degree of the denominator we cannot use the concept of partial fraction then we have to use the long division method to solve the function.