Math Problem Statement

Evaluate the following integral.

Integral from nothing to nothing StartFraction 7 left parenthesis x squared plus 5 right parenthesis Over x left parenthesis x squared plus 7 right parenthesis EndFraction dx∫7x2+5xx2+7 dx

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Part 1

Integral from nothing to nothing StartFraction 7 left parenthesis x squared plus 5 right parenthesis Over x left parenthesis x squared plus 7 right parenthesis EndFraction dx∫7x2+5xx2+7 dxequals=enter your response here

​(Use parentheses to clearly denote the argument of each​ function.)

Solution

To evaluate the integral of the function

7(x2+5)x(x2+7)dx,\int \frac{7(x^2 + 5)}{x(x^2 + 7)} \, dx,

we can first attempt to simplify the expression or decompose it into simpler terms using algebraic manipulation (such as partial fraction decomposition).

Step 1: Simplify the expression

We can rewrite the integrand:

7(x2+5)x(x2+7)=7x2+35x(x2+7).\frac{7(x^2 + 5)}{x(x^2 + 7)} = \frac{7x^2 + 35}{x(x^2 + 7)}.

Now split this into two separate terms:

7x2+35x(x2+7)=7x2x(x2+7)+35x(x2+7).\frac{7x^2 + 35}{x(x^2 + 7)} = \frac{7x^2}{x(x^2 + 7)} + \frac{35}{x(x^2 + 7)}.

Simplifying each term:

=7x2+7+35x(x2+7).= \frac{7}{x^2 + 7} + \frac{35}{x(x^2 + 7)}.

Step 2: Integrate each term separately

Now, let's evaluate the integrals of each term.

First term: 7x2+7dx\int \frac{7}{x^2 + 7} \, dx

This is a standard arctangent form:

1x2+a2dx=1aarctan(xa).\int \frac{1}{x^2 + a^2} \, dx = \frac{1}{a} \arctan\left(\frac{x}{a}\right).

So, we have:

7x2+7dx=77arctan(x7).\int \frac{7}{x^2 + 7} \, dx = \frac{7}{\sqrt{7}} \arctan\left(\frac{x}{\sqrt{7}}\right).

Second term: 35x(x2+7)dx\int \frac{35}{x(x^2 + 7)} \, dx

We use partial fraction decomposition for this term. Let’s write:

35x(x2+7)=Ax+Bx+Cx2+7.\frac{35}{x(x^2 + 7)} = \frac{A}{x} + \frac{Bx + C}{x^2 + 7}.

Multiply both sides by x(x2+7)x(x^2 + 7) to get:

35=A(x2+7)+(Bx+C)x.35 = A(x^2 + 7) + (Bx + C)x.

Expanding:

35=Ax2+7A+Bx2+Cx.35 = A x^2 + 7A + Bx^2 + Cx.

Now, equate the coefficients from both sides:

A+B=0(coefficient of x2)A + B = 0 \quad \text{(coefficient of } x^2 \text{)} C=0(coefficient of x)C = 0 \quad \text{(coefficient of } x \text{)} 7A=35(constant term).7A = 35 \quad \text{(constant term)}.

From 7A=357A = 35, we get A=5A = 5, and from A+B=0A + B = 0, we get B=5B = -5. Thus, we can rewrite the decomposition as:

35x(x2+7)=5x5xx2+7.\frac{35}{x(x^2 + 7)} = \frac{5}{x} - \frac{5x}{x^2 + 7}.

Now, integrate each term:

5xdx=5lnx,\int \frac{5}{x} \, dx = 5 \ln|x|, 5xx2+7dx=52ln(x2+7).\int \frac{-5x}{x^2 + 7} \, dx = -\frac{5}{2} \ln(x^2 + 7).

Step 3: Combine all terms

Now, combine all the integrals:

7(x2+5)x(x2+7)dx=77arctan(x7)+5lnx52ln(x2+7)+C.\int \frac{7(x^2 + 5)}{x(x^2 + 7)} \, dx = \frac{7}{\sqrt{7}} \arctan\left(\frac{x}{\sqrt{7}}\right) + 5 \ln|x| - \frac{5}{2} \ln(x^2 + 7) + C.

Final Answer:

7(x2+5)x(x2+7)dx=77arctan(x7)+5lnx52ln(x2+7)+C.\int \frac{7(x^2 + 5)}{x(x^2 + 7)} \, dx = \frac{7}{\sqrt{7}} \arctan\left(\frac{x}{\sqrt{7}}\right) + 5 \ln|x| - \frac{5}{2} \ln(x^2 + 7) + C.

Would you like any further explanation or have other questions about this? Here's a set of related questions for deeper understanding:

  1. What is the process of partial fraction decomposition?
  2. How do you identify the arctangent integral form?
  3. Can you explain the general method for handling integrals involving x(x2+a2)x(x^2 + a^2) in the denominator?
  4. How do we integrate functions of the form 1x(x2+a2)\frac{1}{x(x^2 + a^2)}?
  5. What would happen if the denominator had different powers of xx?

Tip: Always try to simplify the integrand into easier terms (like partial fractions or simple forms) before performing the integration!

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Math Problem Analysis

Mathematical Concepts

Integration
Partial Fraction Decomposition
Arctangent Form
Logarithmic Integrals

Formulas

∫ 1/(x^2 + a^2) dx = (1/a) arctan(x/a)
∫ 1/x dx = ln|x|
Partial Fraction Decomposition: A/(x) + (Bx + C)/(x^2 + a^2)

Theorems

Integration by Partial Fractions
Standard Integral Forms

Suitable Grade Level

Grades 11-12