Calculate the integral: integral from - pi/4 to 0

Calculate the integral: $\int_{-\frac{\pi}{4}}^0 \frac{3x^4+3x^2+1}{x^2+1} \, dx$.

We will simplify the integrand by performing polynomial division, then integrate.

Step 1: Simplify the integrand using polynomial division

We need to divide $3x^4+3x^2+1$ by $x^2+1$.

First division: $\frac{3x^4}{x^2} = 3x^2$, and $3x^2(x^2+1) = 3x^4 + 3x^2$.

Subtract from the numerator:

\[(3x^4+3x^2+1) - (3x^4+3x^2) = 1\]

Therefore:

\[\frac{3x^4+3x^2+1}{x^2+1} = 3x^2 + \frac{1}{x^2+1}\]

Step 2: Rewrite the integral

\[\int_{-\frac{\pi}{4}}^0 \frac{3x^4+3x^2+1}{x^2+1} \, dx = \int_{-\frac{\pi}{4}}^0 \left(3x^2 + \frac{1}{x^2+1}\right) dx\]

Step 3: Integrate term by term

\[= \int_{-\frac{\pi}{4}}^0 3x^2 \, dx + \int_{-\frac{\pi}{4}}^0 \frac{1}{x^2+1} \, dx\]

For the first integral:

\[\int 3x^2 \, dx = 3 \cdot \frac{x^3}{3} = x^3\]

For the second integral:

\[\int \frac{1}{x^2+1} \, dx = \arctan(x) + C\]

Step 4: Apply the Newton-Leibniz formula

\[\int_{-\frac{\pi}{4}}^0 \frac{3x^4+3x^2+1}{x^2+1} \, dx = \left[x^3 + \arctan(x)\right]_{-\frac{\pi}{4}}^0\]

Step 5: Evaluate at the upper limit $x = 0$

\[F(0) = 0^3 + \arctan(0) = 0 + 0 = 0\]

Step 6: Evaluate at the lower limit $x = -\frac{\pi}{4}$

\[F\left(-\frac{\pi}{4}\right) = \left(-\frac{\pi}{4}\right)^3 + \arctan\left(-\frac{\pi}{4}\right)\]

The first term is $\left(-\frac{\pi}{4}\right)^3 = -\frac{\pi^3}{64}$.

For the second term, note that $-\frac{\pi}{4} \approx -0.7854$, and we need $\arctan\left(-\frac{\pi}{4}\right)$. Since $\arctan$ is an odd function, $\arctan\left(-\frac{\pi}{4}\right) = -\arctan\left(\frac{\pi}{4}\right)$. Note that $\arctan\left(\frac{\pi}{4}\right)$ (the arctangent \emph{of} the number $\frac{\pi}{4}\approx0.7854$) should not be confused with $\frac{\pi}{4}$ itself; using a calculator, $\arctan\left(\frac{\pi}{4}\right) \approx 0.6658$ radians, so $\arctan\left(-\frac{\pi}{4}\right) \approx -0.6658$.

Therefore:

\[F\left(-\frac{\pi}{4}\right) = -\frac{\pi^3}{64} - \arctan\left(\frac{\pi}{4}\right)\]

Step 7: Calculate the definite integral

\[\int_{-\frac{\pi}{4}}^0 \frac{3x^4+3x^2+1}{x^2+1} \, dx = F(0) - F\left(-\frac{\pi}{4}\right)\]

\[= 0 - \left(-\frac{\pi^3}{64} - \arctan\left(\frac{\pi}{4}\right)\right)\]

\[= \frac{\pi^3}{64} + \arctan\left(\frac{\pi}{4}\right)\]

Numerical approximation:

Using $\pi \approx 3.14159$ and $\arctan\left(\frac{\pi}{4}\right) \approx 0.6658$:

\[\frac{\pi^3}{64} + \arctan\left(\frac{\pi}{4}\right) \approx \frac{31.006}{64} + 0.6658 \approx 0.4845 + 0.6658 \approx 1.1503\]

Final answer:

\[\boxed{\int_{-\frac{\pi}{4}}^0 \frac{3x^4+3x^2+1}{x^2+1} \, dx = \frac{\pi^3}{64} + \arctan\left(\frac{\pi}{4}\right) \approx 1.1503}\]