Math Problem Statement

integral of x^2-x+6/ x^3 +3x

Solution

We want to evaluate the integral:

x2x+6x3+3xdx\int \frac{x^2 - x + 6}{x^3 + 3x} \, dx

Step 1: Simplify the integrand

The denominator can be factored as:

x3+3x=x(x2+3)x^3 + 3x = x(x^2 + 3)

So, the integral becomes:

x2x+6x(x2+3)dx\int \frac{x^2 - x + 6}{x(x^2 + 3)} \, dx

Step 2: Split the fraction

We can split the fraction into simpler parts by dividing each term in the numerator by x(x2+3)x(x^2 + 3):

x2x+6x(x2+3)=x2x(x2+3)xx(x2+3)+6x(x2+3)\frac{x^2 - x + 6}{x(x^2 + 3)} = \frac{x^2}{x(x^2 + 3)} - \frac{x}{x(x^2 + 3)} + \frac{6}{x(x^2 + 3)}

This simplifies to:

x2x(x2+3)=1x2+3,xx(x2+3)=1x2+3,6x(x2+3)\frac{x^2}{x(x^2 + 3)} = \frac{1}{x^2 + 3}, \quad \frac{x}{x(x^2 + 3)} = \frac{1}{x^2 + 3}, \quad \frac{6}{x(x^2 + 3)}

Thus, the integrand is now:

(1x2+3+5x(x2+3))dx\int \left( \frac{1}{x^2 + 3} + \frac{5}{x(x^2 + 3)} \right) dx

Step 3: Solve the integral

  1. The first part 1x2+3dx\int \frac{1}{x^2 + 3} \, dx is a standard integral:

1x2+3dx=13arctan(x3)\int \frac{1}{x^2 + 3} \, dx = \frac{1}{\sqrt{3}} \arctan\left(\frac{x}{\sqrt{3}}\right)

  1. For the second part 5x(x2+3)dx\int \frac{5}{x(x^2 + 3)} \, dx, we can solve this using a substitution. Let u=x2+3u = x^2 + 3, then du=2xdxdu = 2x \, dx, and this becomes a standard logarithmic integral. After some algebraic steps, we obtain:

5x(x2+3)dx=56lnx2+3+C\int \frac{5}{x(x^2 + 3)} \, dx = \frac{5}{6} \ln|x^2 + 3| + C

Step 4: Combine the results

So, the final answer for the integral is:

x2x+6x3+3xdx=13arctan(x3)+56lnx2+3+C\int \frac{x^2 - x + 6}{x^3 + 3x} \, dx = \frac{1}{\sqrt{3}} \arctan\left(\frac{x}{\sqrt{3}}\right) + \frac{5}{6} \ln|x^2 + 3| + C

Would you like further clarification or a deeper breakdown of any steps?


Here are five related questions to expand your understanding:

  1. How would you approach solving 6x(x2+3)dx\int \frac{6}{x(x^2 + 3)} \, dx separately?
  2. Can you explain the technique of using substitution to simplify integrals?
  3. What is the derivative of arctan(x3)\arctan\left(\frac{x}{\sqrt{3}}\right)?
  4. How does partial fraction decomposition apply to integrals of rational functions?
  5. What other methods can you use to solve complex rational integrals?

Tip: Always check if the denominator can be factored or simplified before attempting an integral—it can make the process much easier!

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Math Problem Analysis

Mathematical Concepts

Integral Calculus
Algebraic Fractions
Substitution in Integration
Logarithmic and Trigonometric Functions

Formulas

Integral of 1 / (x^2 + a^2) dx = (1/a) * arctan(x/a)
Logarithmic integral: ∫ 1 / u du = ln|u| + C
Substitution Method: u = g(x)

Theorems

Arctangent Integration Theorem
Logarithmic Integration Theorem
Substitution Theorem

Suitable Grade Level

Grades 11-12 (Advanced Calculus)