Question:

Let the function $f(x)$ be defined by $f(x) = \frac{1}{x^2 + 4x + 3}$. Suppose $F(x)$ is an antiderivative of $f(x)$.

(a) Find $F(x)$.

(b) Using $F(x)$, evaluate the definite integral:

Answer:

(a) Finding the antiderivative:

To find $F(x)$, we integrate the given function $f(x)$ with respect to $x$. Let's start by factoring the denominator:

Now, we can use partial fractions to break down the expression. Recall that for an expression of the form $\frac{A}{x + h}$, $A$ is a constant that needs to be determined. Thus, we can write:

Multiplying through by the common denominator, we have:

Expanding the equation and collecting like terms:

Since the coefficients of $x$ and the constant term must be equal on both sides of the equation, we can set up a system of equations:

\begin{align*} A+B &= 0 \ 3A + B &= 1 \end{align*}

Solving this system, we find $A = -\frac{1}{2}$ and $B = \frac{1}{2}$. Therefore:

Now, let's integrate $f(x)$:

\begin{align*} F(x) &= \int f(x) ,dx \ &= \int \left(\frac{-\frac{1}{2}}{x+1} + \frac{\frac{1}{2}}{x+3}\right) ,dx \ &= -\frac{1}{2}\int \frac{1}{x+1} ,dx + \frac{1}{2}\int \frac{1}{x+3} ,dx \end{align*}

Using the property of logarithmic functions, we can rewrite this as:

where $C$ is the constant of integration.

(b) Evaluating the definite integral:

To evaluate the definite integral, we substitute the limits of integration into the antiderivative $F(x)$ and subtract the result from the upper limit to the lower limit. Therefore:

\begin{align*} \int_{0}^{1} \frac{1}{x^2 + 4x + 3} ,dx &= F(1) - F(0) \ &= \left(-\frac{1}{2} \ln|1+1| + \frac{1}{2} \ln|1+3| + C\right) - \left(-\frac{1}{2} \ln|0+1| + \frac{1}{2} \ln|0+3| + C\right) \ &= -\frac{1}{2} \ln 2 + \frac{1}{2} \ln 4 - \left(-\frac{1}{2} \ln 1 + \frac{1}{2} \ln 3\right) \ &= -\frac{1}{2} \ln 2 + \frac{1}{2} \ln 4 - \left(0 + \frac{1}{2} \ln 3\right) \ &= -\frac{1}{2} \ln 2 + \frac{1}{2} \ln 4 - \frac{1}{2} \ln 3 \ &= -\frac{1}{2} \ln \left(\frac{2 \cdot 3}{\sqrt{3}}\right) + \frac{1}{2} \ln 4 \ &= -\frac{1}{2} \ln \left(\frac{6}{\sqrt{3}}\right) + \frac{1}{2} \ln 4 \ &= -\frac{1}{2} \ln \left(2\sqrt{2}\right) + \frac{1}{2} \ln 4 \ &= -\frac{1}{2} \left(\ln 2 + \ln\sqrt{2}\right) + \frac{1}{2} \ln 4 \ &= -\frac{1}{2} \ln 2 - \frac{1}{2} \ln 2 + \frac{1}{2} \ln 4 \ &= -\ln 2 + \frac{1}{2} \ln 4 \ &= \ln \sqrt{\frac{4}{2}} \ &= \ln 2 \ &= \boxed{\ln 2} \end{align*}

Therefore, the value of the definite integral $\int_{0}^{1} \frac{1}{x^2 + 4x + 3} \,dx$ is $\ln 2$.

(c) Using basic antiderivatives:

The integral $\int_{0}^{1} \frac{1}{x^2 + 4x + 3} \,dx$ can also be evaluated using basic antiderivatives. To check if this is the case, let's find the antiderivative using basic antiderivative rules.

We observe that the denominator can be factored as $(x+1)(x+3)$, and the numerator is a constant. By applying the substitution $u = x+1$, we have $du = dx$. Now we can rewrite the integral as:

Using partial fractions, we can decompose the integrand as:

Multiplying through by the common denominator, we have:

Solving this equation, we find $A = -\frac{1}{2}$ and $B = \frac{1}{2}$. Therefore:

Substituting back $u = x+1$, we have:

where $C$ is the constant of integration.

Comparing this result with our antiderivative $F(x)$, we see that they are the same except for the constant term. Therefore, we can conclude that the integral $\int_{0}^{1} \frac{1}{x^2 + 4x + 3} \,dx$ can indeed be evaluated using basic antiderivatives.

To find the exact value of the definite integral using basic antiderivatives, we substitute the limits of integration into the antiderivative $-\frac{1}{2} \ln|x+1| + \frac{1}{2} \ln|x+3|$:

\begin{align*} -\frac{1}{2}\ln|1+1| + \frac{1}{2}\ln|1+3| - \left(-\frac{1}{2}\ln|0+1| + \frac{1}{2}\ln|0+3|\right) &= -\frac{1}{2}\ln 2 + \frac{1}{2}\ln 4 - \left(0 + \frac{1}{2}\ln 3\right) \ &= -\frac{1}{2}\ln 2 + \frac{1}{2}\ln 4 - \frac{1}{2}\ln 3 \ &= -\frac{1}{2}\ln\left(\frac{2\cdot3}{\sqrt 3}\right) + \frac{1}{2}\ln 4 \ &= -\frac{1}{2}\ln \sqrt{2} + \frac{1}{2}\ln 4 \ &= -\ln 2 + \frac{1}{2}\ln 4 \ &= \ln \left(\frac{4}{2}\right) \ &= \ln 2 \ &= \boxed{\ln 2} \end{align*}

Therefore, the value of the definite integral $\int_{0}^{1} \frac{1}{x^2 + 4x + 3} \,dx$ using basic antiderivatives is $\ln 2$.