Question

What is the first step in integrating \(\frac{x^{2}+2 x-3}{x+1} ?\)

Short answer

Question: Find the integral of the function \(\frac{x^{2}+2 x-3}{x+1}\). Answer: \(\int \frac{x^{2}+2x-3}{x+1} dx = \frac{1}{2}x^2 + x - 4\ln|x+1| + C\)

Step-by-step solution

Perform Polynomial Long Division

To start, we will perform polynomial long division to simplify the expression. Set up the long division as follows: ``` ____________________ x+1 | x^2 + 2x - 3 ``` Now we perform the long division.

Long Division

Divide the first term of the denominator (x) into the first term of the numerator (x^2), which results in x. Next, multiply the whole divisor (x+1) by x and subtract the result from the numerator. ``` x ____________________ x+1 | x^2 + 2x - 3 -(x^2 + x) ____________________ x - 3 ``` Repeat the process: Divide the first term of the denominator (x) into the new term (x), which results in + 1. ``` x + 1 ____________________ x+1 | x^2 + 2x - 3 -(x^2 + x) ____________________ x - 3 -(x + 1) ____________________ -4 ``` Now we cannot divide x into -4 since -4 has a lower degree than x, so we stop here. The result of the long division is \(x + 1 - \frac{4}{x+1}\).

Integrate the result

Now, with the simplified expression, we can easily integrate term by term: $$\int \frac{x^{2}+2x-3}{x+1} dx = \int (x + 1 - \frac{4}{x+1}) dx$$ $$ = \int x dx + \int 1 dx - 4 \int \frac{1}{x+1} dx$$

Solve the integrals

Now, we solve each integral separately: $$\int x dx = \frac{1}{2}x^{2} + C_1$$ $$\int 1 dx = x + C_2$$ $$4 \int \frac{1}{x+1} dx = 4\ln|x+1|+ C_3$$ Now, we add the results together:

Combine the results

Combine the results of the individual integrals: $$\int \frac{x^{2}+2x-3}{x+1} dx = \frac{1}{2}x^{2} + x + C_1 + C_2 - 4\ln|x+1|+ C_3$$ Since \(C_1\), \(C_2\), and \(C_3\) are constants, we can combine them into a single constant, C: $$ = \frac{1}{2}x^2 + x - 4\ln|x+1| + C$$ So, the integral of the given function can be written as: $$ \int \frac{x^{2}+2x-3}{x+1} dx = \frac{1}{2}x^2 + x - 4\ln|x+1| + C$$

Key concepts

Integration by Parts

Integration by Parts is a useful technique in calculus that often helps when dealing with products of functions. It's derived from the product rule for differentiation. The formula for Integration by Parts is:\[\int u\, dv = uv - \int v\, du\]Here's how it works:

• Choose which part of the integrand will be \(u\) and which will be \(dv\).
• Differentiate \(u\) to get \(du\), and integrate \(dv\) to get \(v\).
• Substitute these into the formula.
• Solve the resulting integrals.

Integration by Parts is particularly useful when one component of the function becomes simpler upon differentiation, such as polynomials, logarithms, or exponential functions.

Logarithmic Integration

Logarithmic Integration is a method typically used when integrating functions that result in a logarithmic expression, often found with rational functions. Specifically, integrals that have the format \( \int \frac{1}{x} dx \) result in \( \ln |x| + C \). This method also applies to integrals of the form \( \int \frac{1}{x+a} dx \), producing the result \( \ln |x+a| + C \). When solving these integrals, just be sure to:

• Recognize when a function can be simplified to the form \( \frac{1}{x} \) or \( \frac{1}{x+a} \).
• Apply the transformation to achieve a logarithmic result.
• Always include the absolute value in the logarithmic function to accommodate both positive and negative values of \(x\).

Logarithmic integration is straightforward but crucial for handling certain rational function integrals.

Polynomial Division

Polynomial Division is an arithmetic process similar to division of numbers, allowing one to divide a polynomial by another polynomial to simplify expressions. Elaborating on division of polynomials:

• Setup: Align the divisor polynomial outside the division bracket with the dividend polynomial inside.


• Division: Begin by dividing the leading term of the dividend by the leading term of the divisor to get the first term of the quotient.


• Multiplication and Subtraction: Multiply the entire divisor by the obtained quotient term and subtract the result from the dividend.


• Repeat: Use the result (or remainder) from subtraction as a new dividend and repeat the process for terms of lower degree.

In the exercise problem, this method is employed to divide \(x^2 + 2x - 3\) by \(x + 1\), simplifying the expression into a mixture of polynomial and rational expressions, facilitating easier integration.

Rational Functions

Rational Functions are expressions formed by the ratio of two polynomials, such as \( \frac{P(x)}{Q(x)} \), where both \(P(x)\) and \(Q(x)\) are polynomials and \(Q(x) eq 0\). They are crucial and often encountered in calculus and algebra.Here's why understanding Rational Functions is important for integration:

• They often require simplification before integration, achieved through techniques like Polynomial Division.


• Integration can involve transformation into logarithmic or partial fraction forms.


• Studying them enhances comprehension of behavior and properties of more complex functions.

In the context of the given exercise, polynomial division turns the rational function \( \frac{x^2 + 2x - 3}{x + 1} \) into simpler terms \( x + 1 - \frac{4}{x+1} \), combining polynomials and logarithms for integrative computation.