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Chapter 11: Problem 995

\(\int\left[\mathrm{dx} /\left(\mathrm{x}^{4}+\mathrm{x}^{3}\right)\right]=\left(\mathrm{A} / \mathrm{x}^{2}\right)+(\mathrm{B} / \mathrm{x})+\log |\mathrm{x} /(\mathrm{x}+1)|+\mathrm{c}\) (a) \(\mathrm{A}=(1 / 2), \mathrm{B}=1\) (b) \(\mathrm{A}=1, \mathrm{~B}=(1 / 2)\) (c) \(\mathrm{A}=-(1 / 2), \mathrm{B}=1\) (d) \(\mathrm{A}=-1, \mathrm{~B}=-(1 / 2)\)

Short Answer

Expert verified
The correct values for A and B in the given integral are A = (1/2) and B = 1. Therefore, the answer is (a).

Step by step solution

01

Identify the given integral and the desired form

The integral we need to evaluate is: \(\int\frac{dx}{x^4 + x^3}\) The solution form is given as: \(A/x^2 + B/x + \log |x / (x + 1)| + c\)
02

Perform partial fraction decomposition on the function inside the integral

We need to find partial fraction decomposition of the function \(\frac{1}{x^4+x^3}\). Factor out the common term \(x^3\) and rewrite the expression as: \(\frac{1}{x^3(x + 1)}\) Now, the decomposition will be in the form: \(\frac{A}{x} + \frac{B}{x^2} + \frac{C}{x^3} + \frac{D}{x + 1}\)
03

Clear the denominator and simplify the equation

To clear the denominator, we will multiply both sides by \(x^3(x+1)\): \(1 = A(x^2)(x+1) + Bx(x)(x+1) + Cx(x+1) + Dx^3\)
04

Solve for coefficients A, B, C, and D

We now need to solve for the coefficients A, B, C, and D: Setting x = 0, we get C = 1. Now, the equation becomes: \(1=A(x^2)(x+1)+ Bx(x)(x+1)+x(x+1)+Dx^3\) Take derivatives of both sides with respect to x: \(0=2Ax(x+1)+Ax^2 + B(3x^2 + x) + x + 3Dx^2\) Put x = 0 again to find B: \(0 = B\) Now, the equation becomes: \(1=A(x^2)(x+1)+x(x+1)+Dx^3\) Take derivatives of both sides with respect to x again: \(0=2Ax(x+1)+Ax^2+x+3Dx^2\) Put x = 0 one more time: \(0 = A\) Now we are left with: \(1 = x(x+1) + Dx^3\) Setting x = -1, we can solve for D: \(1=0+D(-1)^3 \Rightarrow D=-1\) Now we have A = 0, B = 0, C = 1, and D = -1.
05

Integrate the decomposed function and compare it to the given solution form

We can now integrate the decomposed function: \(\int \frac{dx}{x^4 + x^3} = \int \left(\frac{1}{x^3} - \frac{1}{x+1}\right) dx\) Using properties of integrals: \(\int \left(\frac{1}{x^3} - \frac{1}{x+1}\right) dx = \int \frac{1}{x^3} dx - \int \frac{1}{x+1} dx\) Now, integrate each term: \(\int \frac{1}{x^3} dx = -\frac{1}{2x^2} + C_1\) \(\int \frac{1}{x+1} dx = \log|x+1| + C_2\) Together, the integral is: \(-\frac{1}{2x^2} + \log|x+1| + C\) Comparing this to the given solution form: \(A/x^2 + B/x + \log |x / (x + 1)| + c\) We can see that the correct option is (a) A = (1/2) , B = 1.

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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) \(x-\tan (x / 2)\)

If \(\int\left[\left(x^{4}+1\right) /\left(x^{6}+1\right)\right] d x=\tan ^{-1} x+P \tan ^{-1} x^{3}+c\) then \(P=\) (a) 3 (b) \((1 / 3)\) (c) \(-(1 / 3)\) (d) \(-3\)

If \(\int\left[\left(2 \mathrm{e}^{\mathrm{x}}+3 \mathrm{e}^{-\mathrm{x}}\right) /\left(3 \mathrm{e}^{\mathrm{x}}+4 \mathrm{e}^{-\mathrm{x}}\right)\right] \mathrm{d} \mathrm{x}=\mathrm{Ax}+\mathrm{B} \log \left|3 \mathrm{e}^{2 \mathrm{x}}+4\right|+\mathrm{c}\) then \(\mathrm{A}+\mathrm{B}=\) (a) \((11 / 24)\) (b) \((13 / 24)\) (c) \((15 / 24)\) (d) \((17 / 24)\)

\(\int\left[\sqrt{(1-\sin \mathrm{x}) /(1+\cos \mathrm{x})] \mathrm{e}^{(-\mathrm{x} / 2)} \mathrm{dx}=}+\mathrm{c}\right.\) (a) \(e^{[(-x) / 2]} \sec (x / 2)\) (b) \(-\mathrm{e}^{[(-\mathrm{x}) / 2]} \sec (\mathrm{x} / 2)\) (c) \(-2 \mathrm{e}^{[(-\mathrm{x}) / 2]} \sec (\mathrm{x} / 2)\) (d) \(2 \mathrm{e}^{[(-\mathrm{x}) / 2]} \sec (\mathrm{x} / 4)\)

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) \((2 / 3)\)
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