Question

$\int \frac{(x^4+x^2+1)}{(x^2-x+1)}dx=?$

• A. $\frac{x^3}{3}-\frac{x^2}{2}+x+c$
• B. $\frac{x^3}{3}+\frac{x^2}{2}+x+c$
• C. $\frac{x^3}{3}-\frac{x^2}{2}-x+c$
• D. $\frac{x^3}{3}+\frac{x^2}{2}-x+c$

Answer 1

The correct option is B

$\frac{x^3}{3}+\frac{x^2}{2}+x+c$

Explanation for the correct option:

Evaluating the integral:

$\int \frac{(x^4+x^2+1)}{(x^2-x+1)}dx$

$$\begin{gathered} =\int \frac{x^4+2x^2+1-x^2}{(x^2-x+1)}dx \\ =\int \frac{{(x^2+1)}^2-x^2}{(x^2-x+1)}dx[\because {(a+b)}^2=a^2+b^2+2ab] \\ =\int \frac{(x^2+1+x)(x^2+1-x)}{(x^2-x+1)}dx[\because (a^2-b^2)=(a-b\left)\right(a+b\left)\right] \\ =\int (x^2+1+x)dx \\ =\frac{x^3}{3}+x+\frac{x^2}{2}+c[c=\text{constant}] \end{gathered}$$

Hence, Option (B) is the correct answer.