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Chapter 7: Problem 17

Use a table of integrals to determine the following indefinite integrals. $$\int \frac{d x}{\left(16+9 x^{2}\right)^{3 / 2}}$$

Short Answer

Expert verified
Question: Evaluate the following indefinite integral: \(\int \frac{d x}{\left(16+9 x^{2}\right)^{3 / 2}}\) Answer: \(-\frac{1}{9(\sqrt{16+9x^2})}+C\)

Step by step solution

01

Identify the integral type

Observe the given integral \(\int \frac{d x}{\left(16+9 x^{2}\right)^{3 / 2}}\). Notice that it has a form similar to one of the table of integral entries: $$\int \frac{d x}{\sqrt{a^{2}+x^{2}}} = \sinh^{-1}\frac{x}{a}+C$$ In our case, \(a^2 = 16\) and so \(a = 4\). Now, since the given integral has the power of \(3/2\), we need to use substitution to match the exponent in order to use the arcsinh rule.
02

Use substitution to match the exponent

We will make the following substitution: $$u^2 = 16 + 9x^2$$ Differentiating both sides with respect to \(x\), we get $$2u\frac{d u}{d x} = 18x$$ From this, we can express \(\frac{dx}{du}\), which is: $$\frac{d x}{d u}=\frac{2u}{18x}=\frac{u}{9x}$$ Next, we will rewrite the original integral in terms of \(u\) using this substitution.
03

Rewrite the integral in terms of 'u' and solve

We rewrite the original integral as $$\int \frac{d x}{\left(16+9 x^{2}\right)^{3 / 2}}= \int \frac{1}{u^3}\cdot\left(\frac{u}{9x}\right) d u$$ Now, we can use the arcsinh integral rule to solve this integral: $$\int \frac{1}{u^3}\cdot\left(\frac{u}{9x}\right) d u =\frac{1}{9} \int \frac{d u}{u^2}$$ $$= -\frac{1}{9u} + C$$
04

Substitute 'x' back into the integral

Finally, we need to substitute \(u\) back in terms of \(x\). Recall that we had $$u^2 = 16+9x^2$$ We can solve for \(u\) by taking the square root: $$u = \sqrt{16+9x^2}$$ Now, substitute \(u\) back into the integral: $$-\frac{1}{9u} + C = -\frac{1}{9(\sqrt{16+9x^2})} + C$$
05

Final answer

The indefinite integral is given by: $$\int \frac{d x}{\left(16+9 x^{2}\right)^{3 / 2}} = -\frac{1}{9(\sqrt{16+9x^2})}+C$$

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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. $$f(t)=1 \longrightarrow F(s)=\frac{1}{s}$$

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