/*! 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 13 Evaluate the following integrals... [FREE SOLUTION] | 91影视

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Chapter 7: Problem 13

Evaluate the following integrals. $$\int \frac{3}{(x-1)(x+2)} d x$$

Short Answer

Expert verified
Question: Evaluate the integral $$\int \frac{3}{(x-1)(x+2)} dx$$. Answer: $$\ln\left|\frac{x-1}{x+2}\right|+C$$

Step by step solution

01

Perform Partial Fraction Decomposition

First, we perform partial fraction decomposition on the integrand. We can express the integrand as: $$\frac{3}{(x-1)(x+2)} = \frac{A}{x-1} + \frac{B}{x+2}$$ To solve for A and B, we can multiply both sides by the denominator \((x-1)(x+2)\), which gives: $$3 = A(x+2) + B(x-1)$$ Now, we can solve the system by choosing convenient values for x. Selecting x = 1 results in: $$3 = 3A \Rightarrow A = 1$$ Selecting x = -2 results in: $$3 = -3B \Rightarrow B = -1$$ Thus, the integrand can be rewritten as: $$\frac{1}{x-1} - \frac{1}{x+2}$$
02

Integrate term by term

Now, we integrate both terms with respect to x: $$\int \frac{1}{x-1} dx - \int \frac{1}{x+2} dx$$ For both integrals, we can apply the basic rule of integration for the natural logarithm function: $$\int \frac{1}{u} du = \ln |u| + C$$ Applying this rule, we get: $$\int \frac{1}{x-1} dx - \int \frac{1}{x+2} dx = \ln |x-1| - \ln |x+2| + C$$
03

Combine logarithmic expressions

Finally, we combine the two logarithmic expressions: $$\ln |x-1| - \ln |x+2| + C = \ln\left|\frac{x-1}{x+2}\right|+C$$ This is the final answer for the integral.

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

Suppose that a function \(f\) has derivatives of all orders near \(x=0 .\) By the Fundamental Theorem of Calculus, \(f(x)-f(0)=\int_{0}^{x} f^{\prime}(t) d t\) a. Evaluate the integral using integration by parts to show that $$f(x)=f(0)+x f^{\prime}(0)+\int_{0}^{x} f^{\prime \prime}(t)(x-t) d t.$$ b. Show (by observing a pattern or using induction) that integrating by parts \(n\) times gives $$\begin{aligned} f(x)=& f(0)+x f^{\prime}(0)+\frac{1}{2 !} x^{2} f^{\prime \prime}(0)+\cdots+\frac{1}{n !} x^{n} f^{(n)}(0) \\ &+\frac{1}{n !} \int_{0}^{x} f^{(n+1)}(t)(x-t)^{n} d t+\cdots \end{aligned}$$ This expression is called the Taylor series for \(f\) at \(x=0\).

Graph the function \(f(x)=\frac{1}{x \sqrt{x^{2}-36}}\) on its domain. Then find the area of the region \(R_{1}\) bounded by the curve and the \(x\) -axis on \([-12,-12 / \sqrt{3}]\) and the area of the region \(R_{2}\) bounded by the curve and the \(x\) -axis on \([12 / \sqrt{3}, 12] .\) Be sure your results are consistent with the graph.

Find the volume of the described solid of revolution or state that it does not exist. The region bounded by \(f(x)=(x-1)^{-1 / 4}\) and the \(x\) -axis on the interval (1,2] is revolved about the \(x\) -axis.

Use the following three identities to evaluate the given integrals. $$\begin{aligned}&\sin m x \sin n x=\frac{1}{2}[\cos ((m-n) x)-\cos ((m+n) x)]\\\&\sin m x \cos n x=\frac{1}{2}[\sin ((m-n) x)+\sin ((m+n) x)]\\\&\cos m x \cos n x=\frac{1}{2}[\cos ((m-n) x)+\cos ((m+n) x)]\end{aligned}$$ $$\int \cos x \cos 2 x d x$$
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