Question

Evaluate the integral $\int {\frac{{{x^6} + 1}}{{{x^2} + 1}}} dx$

Answer 1

Hint: Here, the given integral can be solved by simplifying the integral first and
then applying the suitable formulae of integrals.

Given,
$\int {\frac{{{x^6} + 1}}{{{x^2} + 1}}} dx \to (1)$
Now, let us consider the numerator ${x^6} + 1$ as we can see it is in the form of ${a^3} + {b^3}$ where $a = {x^2},b = 1$ .
Since, we know ${a^3} + {b^3} = (a + b)({a^2} - ab + {b^2})$.
Now, we can expand ${x^6} + 1$ as
$\Rightarrow {x^6} + 1 = {({x^2})^3} + 1 = ({x^2} + 1)({x^4} - {x^2} + 1)$
Now, equation (1) can be rewritten as follows:
$\Rightarrow \int {\frac{{({x^2} + 1)({x^4} - {x^2} + 1)}}{{{x^2} + 1}}} dx$
Here, ${x^2} + 1$ term gets cancelled and we will be left with
$\Rightarrow \int {({x^4} - {x^2} + 1)dx}$
Applying, integral to each term, we get
$\Rightarrow \int {{x^4}dx - \int {{x^2}dx + \int {1dx} } }$
As we know that $\int {{x^n}dx = \int {\frac{{{x^{n + 1}}}}{{n + 1}} + c,} }$ where ‘c’ is the
constant of integration. So applying the formulae, we get
$\Rightarrow \frac{{{x^{4 + 1}}}}{{4 + 1}} - \frac{{{x^{2 + 1}}}}{{2 + 1}} + x + c[\because \int {dx} = x] \\\ \Rightarrow \frac{{{x^5}}}{5} - \frac{{{x^3}}}{3} + x + c \\\$
Hence, $\int {\frac{{{x^6} + 1}}{{{x^2} + 1}}} dx$ $= \frac{{{x^5}}}{5} - \frac{{{x^3}}}{3} + x + c$
where ‘c ‘is the constant of integration.
Note: The alternate approach for solving this question is by substitution method where $x = \tan t$ and $dx = {\sec ^2}tdt$.