/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 65 Evaluate the following integrals... [FREE SOLUTION] | 91影视

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Chapter 7: Problem 65

Evaluate the following integrals. $$\int_{1 / 2}^{(\sqrt{2}+3) /(2 \sqrt{2})} \frac{d x}{8 x^{2}-8 x+11}$$

Short Answer

Expert verified
$$ The result of the integral is: $$\frac{1}{16(8x^2 - 8x + 11) - 8} \ln{\frac{7}{3}}.$$

Step by step solution

01

Identify the form of the integrand

The given integral is: $$\int_{1 / 2}^{(\sqrt{2}+3) /(2 \sqrt{2})} \frac{dx}{8x^{2}-8x+11}$$ Let's look at the integrand: $$\frac{1}{8x^{2}-8x+11}$$ It seems to be in the form of a rational function of degree 2 in the denominator.
02

Make a substitution to simplify the integrand

In order to simplify this integrand, we can make the substitution: $$u = 8x^2 - 8x + 11.$$ Differentiating both sides with respect to \(x\), we get: $$\frac{du}{dx} = 16x - 8,$$ which means that $$dx = \frac{du}{16x - 8}.$$
03

Perform the substitution and integrate

Now, substitute \(u\) and \(dx\) in the original integral: \begin{align*} \int_{1 / 2}^{(\sqrt{2}+3) /(2 \sqrt{2})} \frac{dx}{8x^{2}-8x+11} &= \int \frac{\frac{du}{16x - 8}}{u} \\ &= \int \frac{du}{u(16x - 8)}. \end{align*} To evaluate this integral, we need to change the limits of integration as well. Using the substitution \(u = 8x^2 - 8x + 11\): - When \(x = \frac{1}{2}\), \(u = 8\left(\frac{1}{2}\right)^2 - 8\left(\frac{1}{2}\right) + 11 = 3\). - When \(x = \frac{\sqrt{2}+3}{2\sqrt{2}}\), \(u = 8\left(\frac{\sqrt{2}+3}{2\sqrt{2}}\right)^2 - 8\left(\frac{\sqrt{2}+3}{2\sqrt{2}}\right) + 11 = 7\). Now the integral becomes: $$\int_3^7 \frac{du}{u(16x - 8)}$$
04

Find the definite integral and the final result

As the integral is a function of \(u\), we integrate with respect to \(u\): $$\int_3^7 \frac{du}{u(16x - 8)} = \frac{1}{16x - 8} \int_3^7 \frac{du}{u}$$ The integral of \(\frac{1}{u}\) is \(\ln{|u|}\), so we have: \begin{align*} \frac{1}{16x - 8} \int_3^7 \frac{du}{u} &= \frac{1}{16x - 8} [\ln|u|]_3^7 \\ &= \frac{1}{16x - 8} (\ln|7| - \ln|3|) \\ &= \frac{1}{16x - 8} \ln{\frac{7}{3}}. \end{align*} Now substitute \(u\) back in terms of \(x\): $$\frac{1}{16x - 8} \ln{\frac{7}{3}} = \frac{1}{16(8x^2 - 8x + 11) - 8} \ln{\frac{7}{3}}.$$ This is the final result for the given integral: $$\int_{1 / 2}^{(\sqrt{2}+3) /(2 \sqrt{2})} \frac{d x}{8 x^{2}-8 x+11} = \frac{1}{16(8x^2 - 8x + 11) - 8} \ln{\frac{7}{3}}.$$

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