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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 \(\mathrm{f}(\mathrm{x})=\cos \mathrm{x}-\cos ^{2} \mathrm{x}+\cos ^{3} \mathrm{x}-\cos ^{4} \mathrm{x}+\underline{\mathrm{x}}+\) then \(\int f(x) d x=\) \(+c\) (a) \(\tan (\mathrm{x} / 2)\) (b) \(x+\tan (x / 2)\) (b) \(x-(1 / 2) \tan (x / 2)\) (d) \(x-\tan (x / 2)\)

\(\int\left[\left(x^{4}+1\right) /\left\\{x\left(x^{2}+1\right)^{2}\right\\}\right] d x=A \log |x|+\left[B /\left(1+x^{2}\right)\right]+c\) then \(\mathrm{A}=\underline{\mathrm{B}}=\) (a) \(\mathrm{A}=1 ; \overline{\mathrm{B}=-1}\) (b) \(\mathrm{A}=-1 ; \mathrm{B}=1\) (c) \(\mathrm{A}=1 ; \mathrm{B}=1\) (d) \(A=-1 ; B=-1\)

If \(\int[\mathrm{dx} /(5+4 \cos \mathrm{x})]=\operatorname{Ptan}^{-1}[\\{\tan (\mathrm{x} / 2)\\} / 3]+\mathrm{c}\) then \(\mathrm{P}\) (a) \((3 / 2)\) (b) \((1 / 2)\) (c) \((1 / 3)\) (d) \((2 / 3)\)

\(I=\int[1 /\\{(x-4) \sqrt{(x-2)\\}}] d x=f(x)+c\) then \(f(x)\) is (a) \((1 / \sqrt{2}) \log |[\\{\sqrt{(x-2)}-\sqrt{2}\\} /\\{\sqrt{(x-2)}+\sqrt{2}\\}]|+c\) (b) \((1 / 2) \log \mid[\\{\sqrt{x}-2)-\sqrt{2}\\} /\\{\sqrt{(x-2)}+\sqrt{2}\\}] \mid+c\) (c) \((1 / 2 \sqrt{2}) \log \mid[\\{\sqrt{(x-2)}+\sqrt{2}\\} /\\{\sqrt{(x-2)}-\sqrt{2}\\}]+c\) (d) none of these

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