Question

The value of the integral $\int_{0}^{1} \frac{x^3+x+1}{3x^2-3x+4} dx$ is

Answer 1

1/2

Answer 2

To evaluate the integral $\int_{0}^{1} \frac{x^3+x+1}{3x^2-3x+4} dx$, we first perform polynomial long division because the degree of the numerator (3) is greater than the degree of the denominator (2).

Step 1: Polynomial Long Division

Divide $x^3+x+1$ by $3x^2-3x+4$:

So, we can write the integrand as: $\frac{x^3+x+1}{3x^2-3x+4} = \frac{1}{3}x + \frac{1}{3} + \frac{\frac{2}{3}x - \frac{1}{3}}{3x^2-3x+4}$ $= \frac{1}{3}x + \frac{1}{3} + \frac{2x-1}{3(3x^2-3x+4)}$

Step 2: Split the Integral

Now, we can split the integral into two parts: $I = \int_{0}^{1} \left( \frac{1}{3}x + \frac{1}{3} \right) dx + \int_{0}^{1} \frac{2x-1}{3(3x^2-3x+4)} dx$ Let $I_1 = \int_{0}^{1} \left( \frac{1}{3}x + \frac{1}{3} \right) dx$ and $I_2 = \int_{0}^{1} \frac{2x-1}{3(3x^2-3x+4)} dx$.

Step 3: Evaluate $I_1$

$I_1 = \left[ \frac{1}{3} \frac{x^2}{2} + \frac{1}{3} x \right]_{0}^{1}$ $I_1 = \left[ \frac{x^2}{6} + \frac{x}{3} \right]_{0}^{1}$ $I_1 = \left( \frac{1^2}{6} + \frac{1}{3} \right) - \left( \frac{0^2}{6} + \frac{0}{3} \right)$ $I_1 = \frac{1}{6} + \frac{2}{6} - 0 = \frac{3}{6} = \frac{1}{2}$

Step 4: Evaluate $I_2$

$I_2 = \frac{1}{3} \int_{0}^{1} \frac{2x-1}{3x^2-3x+4} dx$ Let $u = 3x^2-3x+4$. Then, $du = (6x-3) dx = 3(2x-1) dx$. So, $(2x-1) dx = \frac{1}{3} du$.

Now, change the limits of integration for $u$: When $x=0$, $u = 3(0)^2 - 3(0) + 4 = 4$. When $x=1$, $u = 3(1)^2 - 3(1) + 4 = 3 - 3 + 4 = 4$.

Substitute $u$ and $du$ into the integral $I_2$: $I_2 = \frac{1}{3} \int_{4}^{4} \frac{\frac{1}{3} du}{u}$ $I_2 = \frac{1}{9} \int_{4}^{4} \frac{1}{u} du$ Since the upper and lower limits of integration are the same, the value of the definite integral is 0. $I_2 = 0$.

Step 5: Combine the Results

The total integral $I = I_1 + I_2$. $I = \frac{1}{2} + 0 = \frac{1}{2}$.

The final answer is $\frac{1}{2}$.

Explanation of the solution:

The integral is evaluated by first performing polynomial long division to simplify the integrand. This splits the integral into two parts: a simple polynomial and a rational function. The polynomial part is integrated directly. The rational function part is integrated using a substitution where the numerator is a multiple of the derivative of the denominator. A key observation is that the limits of integration for the substituted variable become identical, making the second part of the integral zero.

Answer:

The value of the integral is $\frac{1}{2}$.