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Chapter 7: Problem 9

Use a table of integrals to determine the following indefinite integrals. $$\int \frac{3 u}{2 u+7} d u$$

Short Answer

Expert verified
Question: Determine the indefinite integral of the function $\int \frac{3u}{2u+7}du$. Answer: $$\int \frac{3u}{2u+7}du = \frac{3}{4}\left((2u+7) - 7\ln|2u+7|\right) + C$$

Step by step solution

01

Identify the substitution

Let's identify the substitution that will make this integral easier to solve. Observing the integral, we can see that the denominator consists of \((2u+7)\). We can make a substitution for this term. Let: $$v = 2u + 7$$ Now, differentiate both sides with respect to \(u\): $$\frac{dv}{du} = 2$$ Next, we need to find the differential \(du\) in terms of \(dv\), so multiply both sides by \(du\): $$dv = 2du$$ Now, we can solve for \(du\) by dividing both sides by 2: $$\frac{1}{2}dv = du$$
02

Apply the substitution

Now, we rewrite the integral by replacing \((2u+7)\) with \(v\) and \(du\) with \(\frac{1}{2}dv\): $$\int \frac{3u}{v}\cdot \frac{1}{2}dv$$ We also need to express \(u\) in terms of \(v\). From the substitution \(v=2u+7\), we can solve for \(u\): $$u = \frac{v-7}{2}$$ Substitute this expression for u in the integral: $$\int \frac{3\left(\frac{v-7}{2}\right)}{v}\cdot \frac{1}{2}dv$$
03

Simplify the integral

Now let's simplify the integral: $$\frac{3}{4}\int \frac{v-7}{v}dv$$ Split the fraction into two separate parts: $$\frac{3}{4}\int \left(\frac{v}{v}-\frac{7}{v}\right)dv$$ Simplify the expression: $$\frac{3}{4}\int(1-\frac{7}{v})dv$$
04

Integrate

Now we can integrate term-by-term: $$\frac{3}{4}\int(1-\frac{7}{v})dv = \frac{3}{4}\left(\int 1\, dv - \int \frac{7}{v}dv\right)$$ Using the integral table, we find that: $$\int 1\,dv = v$$ and $$\int \frac{1}{v}dv = \ln|v|$$ So, $$\frac{3}{4}\left(\int 1\, dv - \int \frac{7}{v}dv\right) = \frac{3}{4}\left(v - 7\ln|v|\right) + C$$ where C is the constant of integration.
05

Replace 'v' with the original terms

Finally, replace \(v\) with the original substitution \(v=2u+7\): $$\frac{3}{4}\left((2u+7) - 7\ln|2u+7|\right) + C$$ And this is the final indefinite integral: $$\int \frac{3u}{2u+7}du = \frac{3}{4}\left((2u+7) - 7\ln|2u+7|\right) + C$$

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Most popular questions from this chapter

Determine whether the following statements are true and give an explanation or counterexample. a. To evaluate \(\int \frac{4 x^{6}}{x^{4}+3 x^{2}} d x\), the first step is to find the partial fraction decomposition of the integrand. b. The easiest way to evaluate \(\int \frac{6 x+1}{3 x^{2}+x} d x\) is with a partial fraction decomposition of the integrand. c. The rational function \(f(x)=\frac{1}{x^{2}-13 x+42}\) has an irreducible quadratic denominator. d. The rational function \(f(x)=\frac{1}{x^{2}-13 x+43}\) has an irreducible quadratic denominator.

For what values of \(p\) does the integral \(\int_{2}^{\infty} \frac{d x}{x \ln ^{p} x}\) exist and what is its value (in terms of \(p\) )?

\(A\) 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. $$f(t)=\cos a t \longrightarrow F(s)=\frac{s}{s^{2}+a^{2}}$$

By reduction formula 4 in Section 3 $$\int \sec ^{3} u d u=\frac{1}{2}(\sec u \tan u+\ln |\sec u+\tan u|)+C$$ Graph the following functions and find the area under the curve on the given interval. $$f(x)=\left(x^{2}-25\right)^{1 / 2},[5,10]$$

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