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Chapter 7: Problem 58

Evaluate the following integrals. $$\int \frac{d x}{x^{2}-2 x+10}$$

Short Answer

Expert verified
Question: Evaluate the integral of the given function: $$\int\frac{dx}{x^2 - 2x + 10}$$ Solution: The integral of the given function is $$\frac{1}{3}\arctan\left(\frac{x-1}{3}\right) + C$$.

Step by step solution

01

Complete the square in the denominator

We need to rewrite the denominator as a perfect square. We can do this by adding and subtracting \((\frac{1}{2} \cdot 2)^2 =1\) inside the parenthesis to get: $$x^2 - 2x + 10 = (x^2 - 2x + 1) + 9 = (x-1)^2 + 9$$ Now the integral becomes: $$\int \frac{dx}{(x-1)^2 +9}$$
02

Apply substitution

Now, we'll apply the substitution method using the tangent function. Let's use the following substitution to transform the integral: $$x - 1 = 3\tan\theta$$ So, \(dx = 3\sec^2\theta d\theta\) and we can rewrite the integral as: $$\int \frac{3\sec^2\theta d\theta}{(3\tan\theta)^2 + 9}$$
03

Simplify the expression

Now we can simplify the integral expression: $$\int \frac{3\sec^2\theta d\theta}{9\tan^2\theta + 9}$$ Factor out 9 from the denominator: $$\int \frac{3\sec^2\theta d\theta}{9(\tan^2\theta + 1)}$$ Recall the trigonometric identity \(\tan^2\theta + 1 = \sec^2\theta\). Now substitute this in the expression: $$\int \frac{3\sec^2\theta d\theta}{9\sec^2\theta}$$ Cancel out the \(\sec^2\theta\) terms, and rewrite the integral as: $$\int \frac{3}{9} d\theta$$
04

Integrate and back-substitute

Now we can easily integrate the expression: $$\frac{1}{3} \int d\theta = \frac{1}{3}\theta + C$$ Recall the substitution we made earlier: \(x - 1 = 3\tan\theta\). Now we can find the value of \(\theta\): $$\theta = \arctan\left(\frac{x-1}{3}\right)$$ Now we will substitute the value of \(\theta\) back into the integral: $$\frac{1}{3}\arctan\left(\frac{x-1}{3}\right) + C$$ So the final solution is: $$\int\frac{dx}{x^2 - 2x + 10} = \frac{1}{3}\arctan\left(\frac{x-1}{3}\right) + C$$

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Most popular questions from this chapter

An integrand with trigonometric functions in the numerator and denominator can often be converted to a rational integrand using the substitution \(u=\tan (x / 2)\) or \(x=2 \tan ^{-1} u .\) The following relations are used in making this change of variables. $$A: d x=\frac{2}{1+u^{2}} d u \quad B: \sin x=\frac{2 u}{1+u^{2}} \quad C: \cos x=\frac{1-u^{2}}{1+u^{2}}$$ $$\text { Evaluate } \int \frac{d x}{1+\sin x+\cos x}$$

When is the volume finite? Let \(R\) be the region bounded by the graph of \(f(x)=x^{-p}\) and the \(x\) -axis, for \(x \geq 1.\) a. Let \(S\) be the solid generated when \(R\) is revolved about the \(x\) -axis. For what values of \(p\) is the volume of \(S\) finite? b. Let \(S\) be the solid generated when \(R\) is revolved about the \(y\) -axis. For what values of \(p\) is the volume of \(S\) finite?

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Challenge Show that with the change of variables \(u=\sqrt{\tan x}\) the integral \(\int \sqrt{\tan x} d x\) can be converted to an integral amenabl to partial fractions. Evaluate \(\int_{0}^{\pi / 4} \sqrt{\tan x} d x\)

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