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Chapter 7: Problem 30

Evaluate the following integrals. $$\int_{2}^{4} \frac{x^{2}+2}{x-1} d x$$

Short Answer

Expert verified
Answer: The value of the definite integral $\int_{2}^{4} \frac{x^2 + 2}{x - 1} dx$ is $10 + 9\ln(3)$.

Step by step solution

01

Perform Polynomial Long Division

Since the degree of the numerator is greater than the degree of the denominator, we need to perform polynomial long division on the fraction: Dividing \((x^2 + 2)\) by \((x - 1)\), we get: $$ \frac{x^2 +2}{x-1} = x+1 + \frac{3}{x-1} $$ Now, the integral can be written as: $$ \int_{2}^{4} \left(x + 1 + \frac{3}{x - 1}\right) dx $$
02

Calculate the Antiderivative

Now we need to find the antiderivative of the simplified function. This will involve integrating each term separately: $$ \int(x + 1 + \frac{3}{x - 1}) dx = \int xdx + \int 1 dx + 3\int\frac{1}{x-1}dx$ $$ The antiderivative of \(x\) is \(\frac{1}{2}x^2\), the antiderivative of \(1\) is \(x\), and the antiderivative of \(\frac{1}{x - 1}\) is \(\ln |x - 1|\). So the antiderivative of the function is: $$ \frac{1}{2}x^2 + x + 3\ln|x - 1| + C $$where C is the integration constant.
03

Apply the Fundamental Theorem of Calculus

Now we can apply the fundamental theorem of calculus to find the value of the definite integral: $$ \int_{2}^{4} \left(x + 1 + \frac{3}{x - 1}\right) dx = \left(\frac{1}{2}(4)^2 + (4) + 3\ln|(4) - 1|\right) - \left(\frac{1}{2}(2)^2 + (2) + 3\ln|(2) - 1|\right) $$
04

Simplify and Evaluate the Expression

Now, let's simplify and compute the result: $$ \left(\frac{1}{2}(16) + (4) + 3\ln{3}\right) - \left(\frac{1}{2}(4) + (2) + 3\ln{1}\right) $$ $$ (8+4+9\ln3)-(2+2+0) = 10+9\ln3 $$ So, the value of the integral is: $$ \int_{2}^{4} \frac{x^2 + 2}{x - 1} dx = 10 + 9\ln(3) $$

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Most popular questions from this chapter

\(A\) powerful tool in solving problems in engineering and physics is the Laplace transform. Given a function \(f(t),\) the Laplace transform is a new function \(F(s)\) defined by $$ F(s)=\int_{0}^{\infty} e^{-s t} f(t) d t $$ where we assume that s is a positive real number. For example, to find the Laplace transform of \(f(t)=e^{-t},\) the following improper integral is evaluated: $$ F(s)=\int_{0}^{\infty} e^{-s t} e^{-t} d t=\int_{0}^{\infty} e^{-(s+1) t} d t=\frac{1}{s+1} $$ Verify the following Laplace transforms, where a is a real number. $$f(t)=1 \longrightarrow F(s)=\frac{1}{s}$$

The following integrals require a preliminary step such as long division or a change of variables before using partial fractions. Evaluate these integrals. $$\int \frac{d t}{2+e^{-t}}$$

$$\text { Evaluate } \int \frac{d x}{1+\sin x+\cos x} \text { using the }$$ substitution \(x=2 \tan ^{-1} \theta .\) The identities \(\sin x=2 \sin \frac{x}{2} \cos \frac{x}{2}\) and \(\cos x=\cos ^{2} \frac{x}{2}-\sin ^{2} \frac{x}{2}\) are helpful.

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. $$A: d x=\frac{2}{1+u^{2}} d u \quad B: \sin x=\frac{2 u}{1+u^{2}} \quad C: \cos x=\frac{1-u^{2}}{1+u^{2}}$$ $$\text { Evaluate } \int \frac{d x}{1+\sin x}$$

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. $$A: d x=\frac{2}{1+u^{2}} d u \quad B: \sin x=\frac{2 u}{1+u^{2}} \quad C: \cos x=\frac{1-u^{2}}{1+u^{2}}$$ Verify relation \(A\) by differentiating \(x=2 \tan ^{-1} u\). Verify relations \(B\) and \(C\) using a right-triangle diagram and the double-angle formulas $$\sin x=2 \sin \left(\frac{x}{2}\right) \cos \left(\frac{x}{2}\right) \text { and } \cos x=2 \cos ^{2}\left(\frac{x}{2}\right)-1$$
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