Question

Integrate $\left( {\dfrac{{{x^3}}}{{{x^2} + 1}}} \right)$ with respect to x.

Answer 1

Hint:The given question requires us to integrate a function of x with respect to x. Integration gives us a family of curves. Integrals in math are used to find many useful quantities such as areas, volumes, displacement, etc. integral is always found with respect to some variable, which in this case is x.

Complete step by step answer:
The given question requires us to integrate a rational function $\left( {\dfrac{{{x^3}}}{{{x^2} + 1}}} \right)$ in variable x whose degree of numerator is greater than denominator. So, we cannot integrate the function directly.
We can assign a new variable to the denominator of the rational function $\left( {\dfrac{{{x^3}}}{{{x^2} + 1}}} \right)$ .

Let us assume $\left( {{x^2} + 1} \right) = t$ .
Differentiating both sides of the equation, we get,
$ = 2x\left( {dx} \right) = dt$
Shifting $2$ from left side of the equation to right side of the equation to find the value of $x\left( {dx} \right)$, we get,
$ = x\left( {dx} \right) = \dfrac{{dt}}{2}$
So, the integral $\int {\dfrac{{{x^3}\left( {dx} \right)}}{{\left( {{x^2} + 1} \right)}}} $ can be simplified by substituting the value of $x\left( {dx} \right)$ in terms of t as obtained above.
$ = \int {\dfrac{{x\left( {dx} \right)\left( {{x^2}} \right)}}{{\left( {{x^2} + 1} \right)}}} $
Substituting the value of $\left( {{x^2} + 1} \right)$ as t, we get,
$ = \int {\dfrac{{\left( {t - 1} \right)}}{t}} \dfrac{{dt}}{2}$
$ = \dfrac{1}{2}\int {\left( {1 - \dfrac{1}{t}} \right)} \left( {dt} \right)$
$ = \dfrac{1}{2}\left[ {t - {{\log }_e}t} \right] + c$
$ = \dfrac{1}{2}\left[ {\left( {{x^2} + 1} \right) - {{\log }_e}\left( {{x^2} + 1} \right)} \right] + c$
So, $\left[ {\dfrac{1}{2}\left\{ {\left( {{x^2} + 1} \right) - {{\log }_e}\left( {{x^2} + 1} \right)} \right\}} \right]$ is the integral for the given function of x, $\left( {\dfrac{{{x^3}}}{{{x^2} + 1}}} \right)$ .

Note: The indefinite integrals of certain functions may have more than one answer in different forms. However, all these forms are correct and interchangeable into one another. Indefinite integral gives us the family of curves as we don’t know the exact value of the constant.