Determine whether the integral is convergent or divergent. If it is convergent, evaluate the integral.

[tex]\int_2^{\infty} \frac{3}{x^2 + 3x} \, dx[/tex]



Answer :

To determine whether the integral [tex]\(\int_2^{\infty} \frac{3}{x^2 + 3x} \, dx\)[/tex] is convergent or divergent, and evaluate it if it's convergent, follow these steps:

1. Rewrite the Integrand:
Rewrite the integrand to simplify it:

[tex]\[ \frac{3}{x^2 + 3x} = \frac{3}{x(x + 3)} \][/tex]

2. Use Partial Fraction Decomposition:
Decompose the rational function into partial fractions to simplify the integration process:

[tex]\[ \frac{3}{x(x + 3)} = \frac{A}{x} + \frac{B}{x + 3} \][/tex]

To find [tex]\(A\)[/tex] and [tex]\(B\)[/tex], set up the equation:

[tex]\[ 3 = A(x + 3) + Bx \][/tex]

Equate coefficients for like terms:

[tex]\[ 3 = A(x + 3) + Bx \implies 3 = Ax + 3A + Bx \implies 3 = (A + B)x + 3A \][/tex]

From this, we get two equations:

[tex]\[ A + B = 0 \][/tex]
[tex]\[ 3A = 3 \][/tex]

Solving these, we find:

[tex]\[ A = 1 \quad \text{and} \quad B = -1 \][/tex]

Thus, the integrand can be written as:

[tex]\[ \frac{3}{x(x + 3)} = \frac{1}{x} - \frac{1}{x + 3} \][/tex]

3. Set Up the Integral:
Substitute the partial fractions back into the integral:

[tex]\[ \int_2^{\infty} \frac{3}{x^2 + 3x} \, dx = \int_2^{\infty} \left( \frac{1}{x} - \frac{1}{x + 3} \right) \, dx \][/tex]

4. Integrate Term by Term:
Integrate each term separately:

[tex]\[ \int_2^{\infty} \left( \frac{1}{x} - \frac{1}{x + 3} \right) \, dx = \int_2^{\infty} \frac{1}{x} \, dx - \int_2^{\infty} \frac{1}{x + 3} \, dx \][/tex]

Integrate each term:

[tex]\[ \int \frac{1}{x} \, dx = \ln|x| \quad \text{and} \quad \int \frac{1}{x+3} \, dx = \ln|x + 3| \][/tex]

Therefore:

[tex]\[ \int_2^{\infty} \left( \frac{1}{x} - \frac{1}{x + 3} \right) \, dx = \left[ \ln|x| - \ln|x + 3| \right]_2^{\infty} \][/tex]

5. Evaluate the Definite Integral:
Evaluate the limits from [tex]\(2\)[/tex] to [tex]\(\infty\)[/tex]:

[tex]\[ \left[ \ln|x| - \ln|x + 3| \right]_2^{\infty} = \lim_{t \to \infty} \left( \ln|t| - \ln|t + 3| \right) - \left( \ln|2| - \ln|5| \right) \][/tex]

As [tex]\(t \to \infty\)[/tex]:

[tex]\[ \ln|t| - \ln|t + 3| \to 0 \quad \text{because} \quad \ln|t(t+3)^{-1}| \approx \ln(1) = 0 \][/tex]

Therefore:

[tex]\[ \left( \ln|2| - \ln|5| \right) = -\ln\left(\frac{5}{2}\right) \][/tex]

Hence, the integral converges, and its value is:

[tex]\[ -(\ln 2 - \ln 5) = -\ln 2 + \ln 5 = \ln\left(\frac{5}{2}\right) \][/tex]

So, the improper integral [tex]\(\int_2^{\infty} \frac{3}{x^2 + 3x} \, dx\)[/tex] is convergent, and its value is [tex]\(-\ln (2) + \ln (5)\)[/tex].

Answer:\ln \frac{5}{2}

Step-by-step explanation:

1. Simplify the integrand:

\int_{2}^{\infty} \frac{3}{x^2 + 3x} \, dx = \int_{2}^{\infty} \frac{3}{x(x + 3)} \, dx

2. Partial Fraction Decomposition:

Decompose \frac{3}{x(x + 3)} into partial fractions:

\frac{3}{x(x + 3)} = \frac{A}{x} + \frac{B}{x + 3}

Solving for A and B:

3 = A(x + 3) + Bx

Setting x = 0:

\[

3 = A(0 + 3) \implies A = 1

\]

Setting x = -3:

\[

3 = B(-3) \implies B = -1

\]

Therefore:

\frac{3}{x(x + 3)} = \frac{1}{x} - \frac{1}{x + 3}

3. Rewrite the integral:

\int_{2}^{\infty} \frac{3}{x(x + 3)} \, dx = \int_{2}^{\infty} \left( \frac{1}{x} - \frac{1}{x + 3} \right) \, dx

4. Integrate term by term:

\int_{2}^{\infty} \left( \frac{1}{x} - \frac{1}{x + 3} \right) \, dx = \int_{2}^{\infty} \frac{1}{x} \, dx - \int_{2}^{\infty} \frac{1}{x + 3} \, dx

Find the antiderivative of each term:

\int \frac{1}{x} \, dx = \ln|x| + C

\int \frac{1}{x + 3} \, dx = \ln|x + 3| + C

5. Evaluate the definite integrals:

\left[ \ln|x| \right]{2}^{\infty} - \left[ \ln|x + 3| \right]{2}^{\infty}

Consider the limit as x \to \infty:

\lim_{b \to \infty} \left( \int_{2}^{b} \frac{1}{x} \, dx - \int_{2}^{b} \frac{1}{x + 3} \, dx \right)

= \lim_{b \to \infty} \left( \left[ \ln|x| \right]{2}^{b} - \left[ \ln|x + 3| \right]{2}^{b} \right)

= \lim_{b \to \infty} \left( \ln b - \ln 2 \right) - \lim_{b \to \infty} \left( \ln (b + 3) - \ln 5 \right)

6. Simplify the expression:

= \lim_{b \to \infty} \left( \ln b - \ln (b + 3) \right) + (\ln 5 - \ln 2)

= \lim_{b \to \infty} \left( \ln \frac{b}{b + 3} \right) + \ln \frac{5}{2}

As b \to \infty, \frac{b}{b + 3} \to 1, and \ln 1 = 0:

= 0 + \ln \frac{5}{2}

= \ln \frac{5}{2}

Therefore, the integral is convergent, and its value is:

\ln \frac{5}{2}