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

Answer

Verified

363.9k+ views

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

Last updated date: 28th Sep 2023

•

Total views: 363.9k

•

Views today: 5.63k

Recently Updated Pages

What do you mean by public facilities

Difference between hardware and software

Disadvantages of Advertising

10 Advantages and Disadvantages of Plastic

What do you mean by Endemic Species

What is the Botanical Name of Dog , Cat , Turmeric , Mushroom , Palm

Trending doubts

How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Difference Between Plant Cell and Animal Cell

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

The equation xxx + 2 is satisfied when x is equal to class 10 maths CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

Drive an expression for the electric field due to an class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

What is the past tense of read class 10 english CBSE