Question

Integration of 3x^2+9x+1/2x+1

Answer 1

\frac{3x^2}{4} + \frac{15x}{4} - \frac{11}{8} \ln|2x+1| + C

Answer 2

The problem asks for the integration of the expression $\frac{3x^2+9x+1}{2x+1}$.

Since the degree of the numerator is greater than or equal to the degree of the denominator, we first perform polynomial long division.

Step 1: Polynomial Long Division

Divide $3x^2+9x+1$ by $2x+1$.

So, we can write the expression as: $\frac{3x^2+9x+1}{2x+1} = \frac{3}{2}x + \frac{15}{4} - \frac{11/4}{2x+1}$

Step 2: Integration

Now, we integrate each term: $\int \left( \frac{3}{2}x + \frac{15}{4} - \frac{11/4}{2x+1} \right) dx$ We can split this into three separate integrals: $\int \frac{3}{2}x \, dx + \int \frac{15}{4} \, dx - \int \frac{11}{4(2x+1)} \, dx$ Apply the power rule for integration $\int x^n dx = \frac{x^{n+1}}{n+1} + C$ and the rule for $\int \frac{1}{ax+b} dx = \frac{1}{a} \ln|ax+b| + C$.

1. $\int \frac{3}{2}x \, dx = \frac{3}{2} \cdot \frac{x^{1+1}}{1+1} = \frac{3}{2} \cdot \frac{x^2}{2} = \frac{3x^2}{4}$

2. $\int \frac{15}{4} \, dx = \frac{15}{4}x$

3. $\int \frac{11}{4(2x+1)} \, dx = \frac{11}{4} \int \frac{1}{2x+1} \, dx$ Using the formula $\int \frac{1}{ax+b} dx = \frac{1}{a} \ln|ax+b|$, here $a=2$ and $b=1$. So, $\frac{11}{4} \cdot \frac{1}{2} \ln|2x+1| = \frac{11}{8} \ln|2x+1|$

Combining these results and adding the constant of integration $C$: $\int \frac{3x^2+9x+1}{2x+1} \, dx = \frac{3x^2}{4} + \frac{15x}{4} - \frac{11}{8} \ln|2x+1| + C$