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Chapter 11: Problem 974

\(\int[\mathrm{dx} / \sqrt{(x-4)(x-7)}]=+c \quad(4

Short Answer

Expert verified
The short answer is: \[\boxed{2 \sin^{-1}\left(\sqrt{\frac{x - 4}{3}}\right) + C}\]

Step by step solution

01

Identify the integral

Given the integral: \[\int\frac{dx}{\sqrt{(x-4)(x-7)}}\]
02

Perform a substitution

Let \(u = \frac{x - 4}{3}\), which implies \(x = 3u + 4\). Then differentiate x in terms of u: \[\frac{dx}{du} = 3, \quad\Rightarrow dx = 3 du\] Now, substitute the value of x and dx in the integral: \[\int\frac{3 du}{\sqrt{(3u)(3u-3)}}\]
03

Simplify the integrand

The expression becomes: \[\int\frac{3 du}{\sqrt{9u(3u-3)}} = \int\frac{3 du}{\sqrt{9u^2(3-u)}}\] Now, factor out the 9 and take the square root: \[\int\frac{3 du}{3 \sqrt{u(3-u)}}\] Cancel out the 3: \[\int\frac{du}{\sqrt{u(3-u)}}\]
04

Apply an inverse trigonometric function identity

The given integral reminds us of the following identity: \[\frac{d}{dx}\sin^{-1}\left(\frac{x}{a}\right) = \frac{1}{\sqrt{a^2 - x^2}}\] Comparing the integrand, we see that: \[a^2 - x^2 = 3 - u \quad\Rightarrow (1/3)(a^2 - u^2) = (1/3)(3 - u)\] Setting \(a^2 = 3\), we have: \[\int\frac{du}{\sqrt{(1/3)(a^2 - u^2)}}\] Which is similar to the identity, so we can use the inverse sine function.
05

Find the antiderivative

From the identity, we have: \[2 \sin^{-1}\left(\frac{u}{\sqrt{3}}\right) + C\] Now, substitute back for x: \[2 \sin^{-1}\left(\frac{\frac{x - 4}{3}}{\sqrt{3}}\right) + C = 2 \sin^{-1}\left(\sqrt{\frac{x - 4}{3}}\right) + C\] This matches option (a) \[\boxed{2 \sin^{-1}\left(\sqrt{\frac{x - 4}{3}}\right) + C}\]

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Most popular questions from this chapter

If \(\int\left[\left(3^{x}-1\right) /\left(3^{x}+1\right)\right] d x=k \log \left|3^{(x / 2)}+3^{-(x / 2)}\right|+c\) then \(k=\) (a) \(\log _{3} \mathrm{e}\) (b) \(\log _{\mathrm{e}} 3\) (c) \(2 \log _{3} \mathrm{e}\) (d) \(2 \log _{\mathrm{e}} 3\)

\(\int(x+4)(x+3)^{7} d x=\) (a) \(\left[(x+3)^{9} / 9\right]-\left[(x+3)^{8} / 8\right]\) (b) \(\left[\left\\{(x+3)^{8}(8 x+33)\right\\} / 72\right]\) (c) \(\left[\left\\{(x+3)^{8}(8 x+33)\right\\} / 72\right]\) (d) \(\left[(x+3)^{8} / 8\right]\)

\(\int\left[\left(x^{2}+1\right) /\left(x^{4}-x^{2}+1\right)\right] d x=\ldots+c\) (a) \(\tan ^{-1}\left[\left(\mathrm{x}^{2}+1\right) / \mathrm{x}\right]\) (b) \(\tan ^{-1}\left[\left(x^{2}-1\right) / x\right]\) (c) \(\tan ^{-1}(\mathrm{x}+1)\) (d) \(\tan ^{-1}(\mathrm{x}-1)\)

\(\int\left[\left(x^{2} d x\right) /\left\\{\left(x^{2}+2\right)\left(x^{2}+3\right)\right\\}\right]=\ldots\) (a) \(\sqrt{3} \tan ^{-1}(\mathrm{x} / \sqrt{3})+\sqrt{2} \tan ^{-1}(\mathrm{x} / \sqrt{2})\) (b) \(\sqrt{3} \tan ^{-1}(\mathrm{x} / \sqrt{3})-\sqrt{2} \tan ^{-1}(\mathrm{x} / \sqrt{2})\) (c) \(\tan ^{-1}(\mathrm{x} / \sqrt{3})+\sqrt{2} \tan ^{-1}(\mathrm{x} / \sqrt{2})\) (d) \(\tan ^{-1}(\mathrm{x} / \sqrt{3})-\sqrt{2} \tan ^{-1}(\mathrm{x} / \sqrt{2})\)

\(\left.\int \sqrt{[}\left(\sin x-\sin ^{3} x\right) /\left(1-\sin ^{3} x\right)\right] d x=\ldots\) (a) \((2 / 3) \sin ^{-1}\left(\sin ^{3 / 2} x\right)\) (b) \((2 / 3) \sin ^{-1}\left(\cos ^{3 / 2} \mathrm{x}\right)\) (c) \([(-3) / 2] \sin ^{-1}\left(\sin ^{3 / 2} x\right)\) (d) \((3 / 2) \sin ^{-1}\left(\sin ^{3 / 2} x\right)\)
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