Question

Integrate the given expression: $\int {\dfrac{{{x^2} + 3}}{{{x^6}\left( {{x^2} + 1} \right)}}} dx$

Answer 1

: In order to solve this question, first of all we will add and subtract $2{x^2}$ in the numerator. Then we will group the similar terms. After that we will simplify the terms and split numerator and denominators. Again we will add and subtract $2{x^2}$ and repeat the process. and simplify . After that we will use the formulas as $\int {\dfrac{1}{{{x^n}}}dx} = \left( {n - 1} \right){x^{n - 1}} + c$ and $\int {\dfrac{1}{{{x^2} + 1}}} dx = {\tan ^{ - 1}}x + c$ and simplify further to get the required result.

Complete step by step answer:
Let us consider the given integral,
$I = \int {\dfrac{{{x^2} + 3}}{{{x^6}\left( {{x^2} + 1} \right)}}} dx$
Now, we will make changes in the numerator.Adding and subtracting $2{x^2}$ in the numerator,
$\Rightarrow I = \int {\dfrac{{{x^2} + 3 + 2{x^2} - 2{x^2}}}{{{x^6}\left( {{x^2} + 1} \right)}}} dx$
By grouping similar terms, we get
$\Rightarrow I = \int {\dfrac{{3{x^2} + 3 - 2{x^2}}}{{{x^6}\left( {{x^2} + 1} \right)}}} dx$
Taking $3$ common from the first two terms of the numerator,
$\Rightarrow I = \int {\dfrac{{3\left( {{x^2} + 1} \right) - 2{x^2}}}{{{x^6}\left( {{x^2} + 1} \right)}}} dx$
We can split the numerator and the denominator,
$\Rightarrow I = \int {\dfrac{{3\left( {{x^2} + 1} \right)}}{{{x^6}\left( {{x^2} + 1} \right)}} - \dfrac{{2{x^2}}}{{{x^6}\left( {{x^2} + 1} \right)}}} dx$

We can see that the term $\left( {{x^2} + 1} \right)$ in the first part and the ${x^6}$ in the second term gets cancelled.
$\Rightarrow I = \int {\dfrac{3}{{{x^6}}} - \dfrac{2}{{{x^4}\left( {{x^2} + 1} \right)}}} dx$
We can separate the terms and write as,
$\Rightarrow I = \int {\dfrac{3}{{{x^6}}}dx - \int {\dfrac{2}{{{x^4}\left( {{x^2} + 1} \right)}}} } dx$
Let us keep the first term as it is. We are going to add and subtract $2{x^2}$ again in the numerator.
$\Rightarrow I = \int {\dfrac{3}{{{x^6}}}dx - \int {\dfrac{{2 + 2{x^2} - 2{x^2}}}{{{x^4}\left( {{x^2} + 1} \right)}}} } dx$
We group the terms and simplify as,
$\Rightarrow I = \int {\dfrac{3}{{{x^6}}}dx - \int {\dfrac{{2\left( {{x^2} + 1} \right) - 2{x^2}}}{{{x^4}\left( {{x^2} + 1} \right)}}} } dx$

We split the two terms to get,
$\Rightarrow I = \int {\dfrac{3}{{{x^6}}}dx - \int {\dfrac{{2\left( {{x^2} + 1} \right)}}{{{x^4}\left( {{x^2} + 1} \right)}}} } - \dfrac{{2{x^2}}}{{{x^4}\left( {{x^2} + 1} \right)}}dx$
Multiplying the negative sign and applying integration for both the terms,
$\Rightarrow I = \int {\dfrac{3}{{{x^6}}}dx - \int {\dfrac{{2\left( {{x^2} + 1} \right)}}{{{x^4}\left( {{x^2} + 1} \right)}}} } dx + \int {\dfrac{{2{x^2}}}{{{x^4}\left( {{x^2} + 1} \right)}}} dx$
After cancelling the similar terms, we have
$\Rightarrow I = \int {\dfrac{3}{{{x^6}}}dx - \int {\dfrac{2}{{{x^4}}}} } dx + \int {\dfrac{2}{{{x^2}\left( {{x^2} + 1} \right)}}} dx$

We will split the numerator and the denominator in the last term, and get
$\Rightarrow I = \int {\dfrac{3}{{{x^6}}}dx - \int {\dfrac{2}{{{x^4}}}} } dx + \int {\dfrac{2}{{{x^2}}}dx} - \int {\dfrac{2}{{\left( {{x^2} + 1} \right)}}} dx$
Now, we can integrate each term separately. We keep the constants as they are. We know that,
$\int {\dfrac{1}{{{x^n}}}dx} = \left( {n - 1} \right){x^{n - 1}} + c$
$\Rightarrow \int {\dfrac{1}{{{x^2} + 1}}} dx = {\tan ^{ - 1}}x + c$
Using these formulae,
$\therefore I = \dfrac{3}{{5{x^5}}} - \dfrac{2}{{3{x^3}}} + \dfrac{2}{x} - 2{\tan ^{ - 1}}x + c$

Therefore, the integral of $\int {\dfrac{{{x^2} + 3}}{{{x^6}\left( {{x^2} + 1} \right)}}} dx$ is $\dfrac{3}{{5{x^5}}} - \dfrac{2}{{3{x^3}}} + \dfrac{2}{x} - 2{\tan ^{ - 1}}x + c$.

Note: The given expression contains only algebraic components. So, we need to start by making certain changes in the expression since there is no particular formula unless the expression is in the simplest form. We need to check the numerator and denominator. We need to add and subtract a term which helps us reduce it into the simplest form so that we can apply formulas and get the final answer. Like this question has a lot of steps, so remember to go step by step. There are only two formulae involved but a lot of simplification has to be done so that we can apply the formula. Note that the formula for $\int {\dfrac{1}{{{x^n}}}dx}$ and $\int {{x^n}dx}$ are slightly different. So, be thorough with the formulae.