Integral 1 1 X 2 3 2

Integrating the Rational Function (\displaystyle \int \frac{1}{x^{2}+3x+2},dx)

Introduction

When you first encounter a rational function whose denominator is a quadratic polynomial, the instinct is to simplify it before attempting integration. The integral

[ \int \frac{1}{x^{2}+3x+2},dx ]

is a classic example that illustrates the power of algebraic manipulation, partial fractions, and the natural logarithm. Mastering this problem not only strengthens algebraic skills but also prepares you for more advanced techniques such as trigonometric substitutions and integration by parts. In this article we will dissect the integral step by step, explain the underlying theory, and provide a clear, concise solution that you can apply to similar problems.

Step 1 – Factor the Quadratic Denominator

The first move is to factor the quadratic expression in the denominator.

[ x^{2}+3x+2 = (x+1)(x+2) ]

Factoring is possible because the product of the constant terms (1 \times 2 = 2) and the sum of the linear coefficients (1+2 = 3) matches the middle term. Once factored, the integrand becomes a rational function with distinct linear factors:

[ \frac{1}{x^{2}+3x+2} = \frac{1}{(x+1)(x+2)}. ]

Step 2 – Decompose into Partial Fractions

With distinct linear factors, we can express the fraction as a sum of simpler fractions:

[ \frac{1}{(x+1)(x+2)} = \frac{A}{x+1} + \frac{B}{x+2}. ]

To find (A) and (B), multiply both sides by ((x+1)(x+2)):

[ 1 = A(x+2) + B(x+1). ]

Expanding and collecting like terms gives:

[ 1 = (A+B)x + (2A + B). ]

Since the left side has no (x)-term, the coefficient of (x) must be zero:

[ A + B = 0 \quad\Longrightarrow\quad B = -A. ]

The constant term must equal 1:

[ 2A + B = 1 \quad\Longrightarrow\quad 2A - A = 1 \quad\Longrightarrow\quad A = 1. ]

Thus (B = -1). The partial fraction decomposition is:

[ \frac{1}{(x+1)(x+2)} = \frac{1}{x+1} - \frac{1}{x+2}. ]

Step 3 – Integrate Each Term Separately

Now the integral breaks into two elementary integrals:

[ \int \frac{1}{x+1},dx - \int \frac{1}{x+2},dx. ]

Both integrals are of the form (\int \frac{1}{u},du = \ln|u| + C). Applying that rule:

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[ \int \frac{1}{x+1},dx = \ln|x+1| + C_1, ] [ \int \frac{1}{x+2},dx = \ln|x+2| + C_2. ]

Combining them and absorbing the constants into a single constant (C):

[ \int \frac{1}{x^{2}+3x+2},dx = \ln|x+1| - \ln|x+2| + C. ]

Using the logarithm property (\ln a - \ln b = \ln\left(\frac{a}{b}\right)), we can write a more compact result:

[ \boxed{\displaystyle \int \frac{1}{x^{2}+3x+2},dx = \ln\left|\frac{x+1}{,x+2,}\right| + C }. ]

Scientific Explanation – Why Partial Fractions Work

Partial fraction decomposition works because any rational function whose denominator can be factored into linear or irreducible quadratic factors can be expressed as a sum of simpler fractions. Practically speaking, this is rooted in the cover‑up method and the method of equating coefficients, both of which rely on the fact that polynomials form a Euclidean domain. By breaking the integrand into terms that resemble (\frac{1}{u}), we convert the problem into integrating logarithms, which is straightforward.

Common Pitfalls to Avoid

Forgetting to factor the denominator – If you skip factoring, you’ll miss the opportunity to decompose the fraction.
Incorrect algebra when solving for (A) and (B) – Always double‑check the system of equations.
Dropping the absolute value – The logarithm’s domain requires absolute values to handle negative arguments.
Missing the constant of integration – Always add (+C) at the end; it represents an infinite family of antiderivatives.

FAQ

Question	Answer
What if the denominator had repeated factors?	Use the form (\frac{A}{x-a} + \frac{B}{(x-a)^2}) and solve for constants accordingly.
Can I use substitution instead of partial fractions?	For this particular integral, partial fractions is simpler. On the flip side, substitution works better when the denominator is a perfect square or when a trigonometric substitution is natural. So
Why do we use natural logarithms?	The integral of (\frac{1}{x}) is (\ln
Is there a geometric interpretation?	The integral represents the area under the curve (y = \frac{1}{x^{2}+3x+2}). The logarithmic result reflects the growth rate of the reciprocal of a quadratic function.

Conclusion

Integrating (\displaystyle \int \frac{1}{x^{2}+3x+2},dx) is a textbook example that demonstrates how algebraic manipulation, partial fractions, and basic antiderivatives combine to solve a rational integral. By factoring, decomposing, and integrating term by term, we arrive at a concise logarithmic expression that accurately represents the antiderivative. Mastering this technique equips you to tackle a wide array of rational integrals and deepens your understanding of how algebra and calculus intertwine.