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Question: Which of the following is the result of \(\int{\dfrac{{{x}^{4}}+{{x}^{2}}+1}{{{x}^{2}}-x+1}dx}\) . ...

Which of the following is the result of ∫x4+x2+1x2−x+1dx\int{\dfrac{{{x}^{4}}+{{x}^{2}}+1}{{{x}^{2}}-x+1}dx} .
A. 13x2+12x2+x+C\dfrac{1}{3}{{x}^{2}}+\dfrac{1}{2}{{x}^{2}}+x+C
B. 13x2−12x2+x+C\dfrac{1}{3}{{x}^{2}}-\dfrac{1}{2}{{x}^{2}}+x+C
C. 13x2−12x2−x+C\dfrac{1}{3}{{x}^{2}}-\dfrac{1}{2}{{x}^{2}}-x+C
D. None

Explanation

Solution

We have to find the integration of ∫x4+x2+1x2−x+1dx\int{\dfrac{{{x}^{4}}+{{x}^{2}}+1}{{{x}^{2}}-x+1}dx} . Since the degree of the numerator is greater than the denominator, we will perform long division. We will divide x4+x2+1{{x}^{4}}+{{x}^{2}}+1 by x2−x+1{{x}^{2}}-x+1. Hence, we will get x2+x+1{{x}^{2}}+x+1 as the quotient. Now , let us integrate this with respect to xx to get the required value.

Complete step by step answer:
We have to the find value of ∫x4+x2+1x2−x+1dx\int{\dfrac{{{x}^{4}}+{{x}^{2}}+1}{{{x}^{2}}-x+1}dx} . We can see that the degree of the numerator is greater than the denominator. Hence, let us divide x4+x2+1{{x}^{4}}+{{x}^{2}}+1 by x2−x+1{{x}^{2}}-x+1 using long division method.

Hence, by dividing we will get x2+x+1{{x}^{2}}+x+1 .
Now let us integrate this with respect to xx .
∫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
Let us now integrate each of the terms. That is,
∫(x2+x+1)dx=∫x2dx+∫xdx+∫1dx\int{({{x}^{2}}+x+1)}dx=\int{{{x}^{2}}dx}+\int{xdx}+\int{1dx}
We know that ∫xndx=xn+1n+1+C\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}+C and ∫dx=x\int dx=x .
By using the above formulas, let us now integrate. We will get
∫(x2+x+1)dx=x2+12+1+x1+11+1+x+C\int{({{x}^{2}}+x+1)}dx=\dfrac{{{x}^{2+1}}}{2+1}+\dfrac{{{x}^{1+1}}}{1+1}+x+C
Now, by solving this, we will get
∫(x2+x+1)dx=x33+x22+x+C\int{({{x}^{2}}+x+1)}dx=\dfrac{{{x}^{3}}}{3}+\dfrac{{{x}^{2}}}{2}+x+C
The above equation can also be written as
∫(x2+x+1)dx=13x3+12x2+x+C\int{({{x}^{2}}+x+1)}dx=\dfrac{1}{3}{{x}^{3}}+\dfrac{1}{2}{{x}^{2}}+x+C

So, the correct answer is “Option D”.

Note: You must be thorough with standard integration formulas. You may make mistake in ∫xndx=xn+1n+1+C\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}+C, by writing ∫xndx=xn−1n−1+C\int{{{x}^{n}}dx}=\dfrac{{{x}^{n-1}}}{n-1}+C . Also the integral of a constant will be a variable with respect to which the integration is done and not zero like differentiation. Also, when performing long division, do put the coefficient of other lower degree terms as zero, for example, the dividend x4+x2+1{{x}^{4}}+{{x}^{2}}+1 is written as x4+0x3+x2+1{{x}^{4}}+0{{x}^{3}}+{{x}^{2}}+1 . Do not forget to put CC after integration.