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Chapter 1: Problem 20

Evaluate the following integrals: $$ \int \frac{3 x^{3}-8 x+5}{\sqrt{x^{2}-4 x-7}} d x $$

Short Answer

Expert verified
Question: Evaluate the following integral: $$\int \frac{3x^3 - 8x + 5}{\sqrt{x^2 - 4x - 7}}dx$$ Answer: $$\frac{\sqrt{11}(x^3 - 8x + 5)}{33\sqrt{(x-2)^2 - 11}} + C$$

Step by step solution

01

Complete the square in the denominator

We first complete the square for the quadratic expression under the square root sign in the denominator: $$ x^2 - 4x - 7 = (x - 2)^2 - 11 $$ This allows us to rewrite the integral as: $$ \int \frac{3 x^{3}-8 x+5}{\sqrt{(x - 2)^2 - 11}} d x $$
02

Find an appropriate substitution

Notice the integral contains a term with the square root of a quadratic expression in the denominator, which suggests a trigonometric substitution. In this case, let's consider: $$ x - 2 = \sqrt{11} \tan(\theta) $$ Now, differentiate x with respect to 胃 to find the appropriate substitution for dx: $$ \frac{d x}{d \theta} = \sqrt{11} \sec^2(\theta) $$ Thus we have: $$dx = \sqrt{11}\sec^2(\theta)d\theta$$
03

Rewrite and simplify the integral using the substitution

Substitute the expressions for x and dx into the integral: $$ \int \frac{3(2 + \sqrt{11}\tan{(\theta)})^3 - 8(2 + \sqrt{11}\tan{(\theta)}) + 5}{\sqrt{(11(\tan^2(\theta)))}} \sqrt{11}\sec^2(\theta) d\theta $$ After substituting, expand and simplify the integral expression: $$ \int \frac{3(\tan^3(\theta)+6\tan^2(\theta)+12\tan(\theta)+8)\sqrt{11}-8\sqrt{11}\tan(\theta)-16\sqrt{11}+5\sqrt{11}}{11\tan(\theta)} d\theta $$ $$ \int \frac{3\sqrt{11}(\tan^3(\theta)+6\tan^2(\theta)+12\tan(\theta)+8)-16\sqrt{11}\tan(\theta)}{11\tan(\theta)} d\theta $$
04

Separate the integral and integrate

Now, separate the integral into manageable parts: $$ \int \left(\frac{3\sqrt{11}(\tan^3(\theta)+6\tan^2(\theta))}{11\tan(\theta)}+\frac{3\sqrt{11}(12\tan(\theta)+8) - 16\sqrt{11}\tan(\theta)}{11\tan(\theta)}\right) d\theta $$ Integrate each term: $$ \int \frac{3\sqrt{11} \tan^2(\theta)}{11} d\theta + \int \frac{3\sqrt{11} \tan(\theta)}{11} d\theta - \int \frac{16\sqrt{11}}{11} d\theta + \int \frac{3\sqrt{11} \sec^2(\theta)}{11} d\theta $$ Integrating, we get: $$ \frac{3\sqrt{11}}{11\cdot3}(\tan^3(\theta)) + \frac{3\sqrt{11}}{11\cdot2}(\tan^{2}(\theta)) - \frac{16\sqrt{11}}{11}(\theta) + \frac{3\sqrt{11}}{11}(\tan(\theta)) $$
05

Undo substitution and simplify

Undo the trigonometric substitution, replacing 胃 with x: $$ \frac{3\sqrt{11}}{33}(2 + \sqrt{11}\tan{(\theta)})^3 + \frac{3\sqrt{11}}{22}(2+\sqrt{11}\tan{(\theta)})^2 - \frac{16\sqrt{11}}{11}(\arctan(\frac{x-2}{\sqrt{11}})) + \frac{3\sqrt{11}}{11}(2+\sqrt{11}\tan{(\theta)}) $$ Simplify the result: $$ \frac{\sqrt{11}(x^3 - 8x + 5)}{33\sqrt{(x-2)^2 - 11}} + C, $$ where C represents the constant of integration. This is the final result of the given integral: $$ \int \frac{3x^3 - 8x + 5}{\sqrt{x^2 - 4x - 7}}dx = \frac{\sqrt{11}(x^3 - 8x + 5)}{33\sqrt{(x-2)^2 - 11}} + C. $$

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