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Question: The \[\int {\dfrac{{{x^4} + {x^2} + 1}}{{{x^2} - x + 1}}dx = } \] A. \[\dfrac{{{x^3}}}{3} - \dfrac...

The ∫x4+x2+1x2−x+1dx=\int {\dfrac{{{x^4} + {x^2} + 1}}{{{x^2} - x + 1}}dx = }
A. x33−x22+x+c\dfrac{{{x^3}}}{3} - \dfrac{{{x^2}}}{2} + x + c
B. x33+x22+x+c\dfrac{{{x^3}}}{3} + \dfrac{{{x^2}}}{2} + x + c
C. x33−x22−x+c\dfrac{{{x^3}}}{3} - \dfrac{{{x^2}}}{2} - x + c
D. x33+x22−x+c\dfrac{{{x^3}}}{3} + \dfrac{{{x^2}}}{2} - x + c

Explanation

Solution

Firstly, We will write the numerator in terms of perfect square and then convert the numerator in such a way that some of the terms from numerator and denominator get cancelled. Like, we will write x4+x2+1{x^4} + {x^2} + 1 as x4+2x2−x2+1{x^4} + 2{x^2} - {x^2} + 1 and then combine the terms x4+2x2+1{x^4} + 2{x^2} + 1 together to form a perfect square of x2+1{x^2} + 1. So that, we will get numerator to be equal to (x2+1)2−x2{({x^2} + 1)^2} - {x^2} and then use the formula a2−b2=(a+b)(a−b){a^2} - {b^2} = (a + b)(a - b) to write the numerator as a product of two terms. After that, one of the terms gets cancelled with the denominator and then we will Just integrate the remaining terms with respect to xx.

Complete step by step answer:
∫x4+x2+1x2−x+1dx=∫x4+2x2−x2+1x2−x+1dx\int {\dfrac{{{x^4} + {x^2} + 1}}{{{x^2} - x + 1}}dx = \int {\dfrac{{{x^4} + 2{x^2} - {x^2} + 1}}{{{x^2} - x + 1}}dx} }
Clubbing some terms together to convert it into formula (a+b)2=a2+b2+2ab{\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab
∫(x4+2x2+1)−x2x2−x+1dx\int {\dfrac{{({x^4} + 2{x^2} + 1) - {x^2}}}{{{x^2} - x + 1}}dx}
⇒∫(x2+1)2−x2x2−x+1dx\Rightarrow \int {\dfrac{{{{({x^2} + 1)}^2} - {x^2}}}{{{x^2} - x + 1}}dx}
Now, using the formula a2−b2=(a+b)(a−b){a^2} - {b^2} = (a + b)(a - b) in the numerator.
∫(x2+1−x)(x2+1+x)x2−x+1dx\int {\dfrac{{({x^2} + 1 - x)({x^2} + 1 + x)}}{{{x^2} - x + 1}}dx}

Now, cancelling the term x2−x+1{x^2} - x + 1 from numerator and denominator, we get
∫x4+x2+1x2−x+1dx=∫(x2+x+1)dx\int {\dfrac{{{x^4} + {x^2} + 1}}{{{x^2} - x + 1}}dx = \int {({x^2} + x + 1} } )dx
Using ∫(a+b+c)dx=∫adx+∫bdx+∫cdx\int {(a + b + c)} dx = \int a dx + \int b dx + \int c dx, we get
∫x2dx+∫xdx+∫1dx\int {{x^2}} dx + \int x dx + \int 1 dx
Using the formula ∫xndx=xn+1n+1+c\int {{x^n}} dx = \dfrac{{{x^{n + 1}}}}{{n + 1}} + c and ∫1dx=x+c\int 1 dx = x + c for integration, we get
x2+12+1+x1+11+1+x+c\dfrac{{{x^{2 + 1}}}}{{2 + 1}} + \dfrac{{{x^{1 + 1}}}}{{1 + 1}} + x + c
⇒x33+x22+x+c\Rightarrow \dfrac{{{x^3}}}{3} + \dfrac{{{x^2}}}{2} + x + c
Hence, we get
∫x4+x2+1x2−x+1dx=x33+x22+x+c\int {\dfrac{{{x^4} + {x^2} + 1}}{{{x^2} - x + 1}}dx = \dfrac{{{x^3}}}{3} + \dfrac{{{x^2}}}{2} + x + c}

Therefore, the correct option is B.

Note: Whenever the polynomial given in numerator is having greater degree than polynomial in denominator, we try to divide the and simplify the expression such that we arrive at an expression which is easy to integrate. Indefinite integrals can have multiple interconvertible answers as the values of the arbitrary constants can vary.