Question

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

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

$\int x^2dx + \int xdx + \int dx$

• A. $\frac{1}{3}x^3 + \frac{1}{2}x^2 + x + c$
• B. $\frac{1}{3}x^3 - \frac{1}{2}x^2 + x + c$
• C. $\frac{1}{3}x^3 + \frac{1}{2}x^2 - x + c$
• D. None of these

Answer 1

Answer: (A)

A) $\frac{1}{3}x^3 + \frac{1}{2}x^2 + x + c$

Answer 2

The problem asks us to evaluate the integral $\int \frac{x^4 + x^2 + 1}{x^2 - x + 1} dx$.

The key to solving this integral is to simplify the integrand by factoring the numerator.
We use the algebraic identity:
$a^4 + a^2 b^2 + b^4 = (a^2 + ab + b^2)(a^2 - ab + b^2)$

In our case, we can consider $a=x$ and $b=1$.
So, the numerator $x^4 + x^2 + 1$ can be factored as:
$x^4 + x^2(1)^2 + (1)^4 = (x^2 + x(1) + 1^2)(x^2 - x(1) + 1^2)$
$x^4 + x^2 + 1 = (x^2 + x + 1)(x^2 - x + 1)$

Now, substitute this factored form back into the integral: $\int \frac{(x^2 + x + 1)(x^2 - x + 1)}{x^2 - x + 1} dx$

Since $x^2 - x + 1$ has a discriminant $\Delta = (-1)^2 - 4(1)(1) = 1 - 4 = -3 < 0$ and its leading coefficient is positive (1 > 0), $x^2 - x + 1$ is always positive for all real values of $x$. Thus, it is never zero, and we can safely cancel the term $(x^2 - x + 1)$ from the numerator and the denominator.

The integral simplifies to: $\int (x^2 + x + 1) dx$

Now, we can integrate each term separately using the power rule for integration, $\int x^n dx = \frac{x^{n+1}}{n+1} + C$: $\int x^2 dx + \int x dx + \int 1 dx$ $= \frac{x^{2+1}}{2+1} + \frac{x^{1+1}}{1+1} + \frac{x^{0+1}}{0+1} + C$ $= \frac{x^3}{3} + \frac{x^2}{2} + x + C$

Comparing this result with the given options:
A) $\frac{1}{3}x^3 + \frac{1}{2}x^2 + x + c$
B) $\frac{1}{3}x^3 - \frac{1}{2}x^2 + x + c$
C) $\frac{1}{3}x^3 + \frac{1}{2}x^2 - x + c$
D) None of these

The calculated integral matches option A.