Integrate y³+4y²/y³+y dy

First, factor out $y$ from the numerator and denominator:

$$\frac{y^3+4y^2}{y^3+y} = \frac{y^2(y+4)}{y(y^2+1)} = \frac{y(y+4)}{y^2+1} = \frac{y^2+4y}{y^2+1}$$

Since the degree of the numerator is equal to the degree of the denominator, perform polynomial long division or rewrite the numerator:

$$\frac{y^2+4y}{y^2+1} = \frac{(y^2+1) + 4y - 1}{y^2+1} = 1 + \frac{4y-1}{y^2+1}$$

So the integral becomes:

$$\int \left( 1 + \frac{4y-1}{y^2+1} \right) dy$$

The first integral is straightforward:

$$\int 1 \, dy = y$$

For the second integral, $\int \frac{4y}{y^2+1} \, dy$, use a substitution. Let $u = y^2+1$, so $du = 2y \, dy$. Then $4y \, dy = 2 \, du$.

$$\int \frac{4y}{y^2+1} \, dy = \int \frac{2 \, du}{u} = 2 \ln|u| + C_1 = 2 \ln(y^2+1) + C_1$$

(Since $y^2+1$ is always positive, we can remove the absolute value.)
The third integral is a standard arctangent form:

$$\int \frac{1}{y^2+1} \, dy = \arctan(y) + C_2$$

$$\int \frac{y^3+4y^2}{y^3+y} dy = y + 2 \ln(y^2+1) - \arctan(y) + C$$

The final answer is $y + 2 \ln(y^2+1) - \arctan(y) + C$.