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Chapter 23: Problem 641

Integrate the expression: \(\int\left[\\{(2 x+7) d x\\} /\left(x^{2}+2 x+5\right)\right]\).

Short Answer

Expert verified
The short version of the solution is: Complete the square to get: \(\int\frac{(2x+7)dx}{(x+1)^2 + 4}\). Perform a substitution \(x+1 = 2\sin{u}\) and simplify to get: \(\frac{1}{2}\int(4\sin{u}+5)\cos{u} \, du\). Evaluate the integral: \(2\sin^2(u) + \frac{5}{2}\sin(u) + C\). Substitute back the original variable: \((x+1)^2 + \frac{5}{2}(x+1) + C\). Hence, \(\int\frac{(2x+7)dx}{x^2 + 2x + 5} = (x+1)^2 + \frac{5}{2}(x+1) + C\).

Step by step solution

01

Complete the square in the denominator

First, we need to complete the square in the denominator. The given expression is: \(\int\frac{(2x+7)dx}{x^2 + 2x + 5}\) To complete the square, we group the terms with x and find the square of half of the coefficient of x, which will be added and subtracted inside the expression: \(x^2 + 2x = (x^2 + 2x + 1) - 1\) Thus, the denominator becomes: \(x^2 + 2x + 5 = (x^2 + 2x + 1) + 4 = (x+1)^2 + 4\) Now, the integral looks like: \(\int\frac{(2x+7)dx}{(x+1)^2 + 4}\)
02

Perform a substitution

We will perform the substitution \(x+1 = 2\sin{u}\), and consequently, \(dx = 2\cos{u}du\) . This substitution will make our integral look more like a trigonometric function: Now the integral looks like: \(\int\frac{[(2(2\sin{u}-1)+7)(2\cos{u}\, du)]}{[2\sin(u)]^2 + 4}\)
03

Simplify the integral

Let's simplify the integral by substituting the new variable: \(\int\frac{(4\sin{u}+5)(2\cos{u}\, du)}{4\sin^2(u) + 4}\) Now, we can see that the constant 4 can be factored out in the denominator: \(\int\frac{(4\sin{u}+5)(2\cos{u}\, du)}{4(\sin^2(u) + 1)}\) Since we know the trigonometric identity \(\sin^2(u) + \cos^2(u) = 1\), the denominator simplifies to: \(\int\frac{(4\sin{u}+5)(2\cos{u}\, du)}{4}\) Now, the integral is: \(\frac{1}{2}\int(4\sin{u}+5)\cos{u} \, du\)
04

Evaluate the integral

We can now proceed with the integration. Split the integral into two parts and integrate each part with respect to u: \(\frac{1}{2}\int(4\sin{u}+5)\cos{u} \, du = \frac{1}{2}\int(4\sin{u}\cos{u}\, du) + \frac{1}{2}\int(5\cos{u} \,du)\) Now, evaluating both integrals: \(\frac{1}{2}\int(4\sin{u}\cos{u} \, du) = 2\sin^2(u) + C_1\) \(\frac{1}{2}\int(5\cos{u} \, du) = \frac{5}{2}\sin(u) + C_2\) The combined integral becomes: \(2\sin^2(u) + \frac{5}{2}\sin(u) + C\), where \(C = C_1 + C_2\)
05

Substitute back the original variable

Now, remember we made the substitution \(x+1 = 2\sin{u}\), so we will reverse this and get the final answer in terms of x: \(\sin(u) = \frac{1}{2}(x+1)\) The final antiderivative is: \(2\left(\frac{1}{2}(x+1)\right)^2 + \frac{5}{2}(\frac{1}{2}(x+1)) + C\) Simplify: \((x+1)^2 + \frac{5}{2}(x+1) + C\) So, the integral is: \(\int\frac{(2x+7)dx}{x^2 + 2x + 5} = (x+1)^2 + \frac{5}{2}(x+1) + C\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Completing the Square
Completing the square is a helpful technique in integration. When you encounter quadratic expressions, this method converts them into a perfect square form. This makes it easier to handle and understand.

For the expression \(x^2 + 2x + 5\), completing the square rearranges it into \((x+1)^2 + 4\). The steps involve:
  • Identifying the terms \(x^2 + 2x\), representing the initial part of the quadratic.
  • Taking half of the coefficient of \(x\), squaring it, and adding/subtracting this inside the expression. Here, half of 2 is 1, and its square is 1.
  • Rewrite \(x^2 + 2x + 5\) as \((x+1)^2 + 4\) by adjusting appropriately.
This new form is manageable for methods like trigonometric substitution or finding antiderivatives. Always start by trying to complete the square with quadratics.
Trigonometric Substitution
Trigonometric substitution is a powerful technique in integration, especially for expressions involving squares. By swapping variables using trigonometric identities, integration becomes simplified.

In our exercise, we used the substitution \(x+1 = 2\sin{u}\), transforming the integral under a trigonometric identity. This helps tackle expressions like \((x+1)^2 + 4\), converting them into a format that correlates with basic trigonometric identities:
  • Substitute \(x+1\) with \(2\sin{u}\), converting the expression to trigonometric form.
  • Calculate \(dx\) as \(2\cos{u}du\), simplifying the integration process.
  • Recognize the trigonometric identity: \(\sin^2(u) + \cos^2(u) = 1\).
The substitution reduces complexity by leveraging angles and sides relations within triangles. This facilitates easier calculations when finding antiderivatives.
Antiderivatives
Finding antiderivatives is a key part of integration. It involves reversing the differentiation process to obtain the original function before it was differentiated.

In our exercise, you find antiderivatives by integrating term by term after substitution and simplifications. Here鈥檚 a step-by-step of what we did:
  • After substituting, rewrite the integral in simpler terms.
  • Break it down into manageable parts: \(\int (4\sin{u}\cos{u} \, du)\) and \(\int (5\cos{u} \, du)\).
  • Integrate each part separately, using basic trigonometric integrals. For example, \(\int \sin(2u) \, du\) integrates to \(\sin^2(u)\).
Back-substitute the original variable to express the final antiderivative in terms of \(x\). This gives the fully integrated expression, which helps understand the area under curves or accumulated quantities.

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