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

Last updated date: 24th Mar 2023

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Answer

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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 $dx = {\sec ^2}tdt$.

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 $dx = {\sec ^2}tdt$.

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