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(d) ∫2x+1x2−1dx\int \frac{2 x+1}{x^{2}-1} d x
Sep 10, 2024
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Solution by Steps
step 1
To differentiate the function Y2/x Y^2/x with respect to x x , we apply the quotient rule: ddx(Y2/x)=(0)(x)−(Y2)(1)x2=−Y2x2 \frac{d}{dx}(Y^2/x) = \frac{(0)(x) - (Y^2)(1)}{x^2} = -\frac{Y^2}{x^2}
step 2
The condition for the differentiation is x≠0 x \neq 0 and Y=0 Y = 0 . Thus, we have the domain: {x∈R:x≠0} \{x \in \mathbb{R} : x \neq 0\} and \{y \in \mathbb{R} : (y = 0 \text{ and } Y = 0) \text{ or } (y < 0 \text{ and } Y \neq 0)\}
step 3
The function Y2/x Y^2/x is even since it satisfies the property f(−x)=f(x) f(-x) = f(x) for all x x in its domain
step 4
To integrate 2x+1x2−1 dx \frac{2x + 1}{x^2 - 1} \, dx , we can use partial fraction decomposition: ∫2x+1(x−1)(x+1) dx \int \frac{2x + 1}{(x - 1)(x + 1)} \, dx which leads to 12(3log⁡(1−x)+log⁡(x+1))+C \frac{1}{2} (3 \log(1 - x) + \log(x + 1)) + C
step 5
The final result can also be expressed as 32log⁡(1−x)+12log⁡(1+x)+C \frac{3}{2} \log(1 - x) + \frac{1}{2} \log(1 + x) + C or log⁡((1−x)3(1+x))+C \log(\sqrt{(1 - x)^3 (1 + x)}) + C
step 6
The generalized Puiseux series expansion gives us 12log⁡(8(x+1))−3(x+1)4−316(x+1)2−116(x+1)3−3128(x+1)4+O((x+1)5) \frac{1}{2} \log(8 (x + 1)) - \frac{3 (x + 1)}{4} - \frac{3}{16} (x + 1)^2 - \frac{1}{16} (x + 1)^3 - \frac{3}{128} (x + 1)^4 + O((x + 1)^5)
step 7
The Taylor series expansion around x=0 x = 0 yields −x−x2−x33−x42−x55+O(x6) -x - x^2 - \frac{x^3}{3} - \frac{x^4}{2} - \frac{x^5}{5} + O(x^6)
Answer
The integral ∫2x+1x2−1 dx \int \frac{2x + 1}{x^2 - 1} \, dx evaluates to 12(3log⁡(1−x)+log⁡(x+1))+C \frac{1}{2} (3 \log(1 - x) + \log(x + 1)) + C
Key Concept
Differentiation and Integration of Rational Functions
Explanation
The differentiation and integration of rational functions often involve applying rules such as the quotient rule for differentiation and partial fraction decomposition for integration. The results can be expressed in various forms, including logarithmic and series expansions.
solve step by step
To solve the integral ∫2x+1x2−1 dx, \int \frac{2x + 1}{x^2 - 1} \, dx, we can use the method of partial fractions. First, we factor the denominator: x2−1=(x−1)(x+1). x^2 - 1 = (x - 1)(x + 1). Next, we express the integrand as a sum of partial fractions: 2x+1x2−1=Ax−1+Bx+1. \frac{2x + 1}{x^2 - 1} = \frac{A}{x - 1} + \frac{B}{x + 1}. To find AA and BB, we multiply both sides by the denominator x2−1x^2 - 1: 2x+1=A(x+1)+B(x−1). 2x + 1 = A(x + 1) + B(x - 1). Expanding the right side gives: 2x+1=Ax+A+Bx−B=(A+B)x+(A−B). 2x + 1 = Ax + A + Bx - B = (A + B)x + (A - B). Now, we can set up a system of equations by equating coefficients: 1. A+B=2A + B = 2 (coefficient of xx) 2. A−B=1A - B = 1 (constant term) Now, we can solve this system. Adding the two equations: (A+B)+(A−B)=2+1  ⟹  2A=3  ⟹  A=32. (A + B) + (A - B) = 2 + 1 \implies 2A = 3 \implies A = \frac{3}{2}. Substituting AA back into the first equation: 32+B=2  ⟹  B=2−32=12. \frac{3}{2} + B = 2 \implies B = 2 - \frac{3}{2} = \frac{1}{2}. Now we have: 2x+1x2−1=3/2x−1+1/2x+1. \frac{2x + 1}{x^2 - 1} = \frac{3/2}{x - 1} + \frac{1/2}{x + 1}. We can now integrate each term separately: ∫2x+1x2−1 dx=∫(3/2x−1+1/2x+1) dx. \int \frac{2x + 1}{x^2 - 1} \, dx = \int \left( \frac{3/2}{x - 1} + \frac{1/2}{x + 1} \right) \, dx. This gives us: =32∫1x−1 dx+12∫1x+1 dx. = \frac{3}{2} \int \frac{1}{x - 1} \, dx + \frac{1}{2} \int \frac{1}{x + 1} \, dx. The integrals are: =32ln⁡∣x−1∣+12ln⁡∣x+1∣+C, = \frac{3}{2} \ln |x - 1| + \frac{1}{2} \ln |x + 1| + C, where CC is the constant of integration. Combining the logarithms, we can write: =12(3ln⁡∣x−1∣+ln⁡∣x+1∣)+C. = \frac{1}{2} \left( 3 \ln |x - 1| + \ln |x + 1| \right) + C. Thus, the final result is: ∫2x+1x2−1 dx=12(3ln⁡∣x−1∣+ln⁡∣x+1∣)+C. \int \frac{2x + 1}{x^2 - 1} \, dx = \frac{1}{2} \left( 3 \ln |x - 1| + \ln |x + 1| \right) + C.
Solution by Steps
step 1
The first integral is given by ∫1∞ln⁡xxdx \int_{1}^{\infty} \frac{\ln x}{x} dx . This integral does not converge
step 2
The second integral is ∫xcos⁡2xdx \int \frac{x}{\cos^2 x} dx . This can be simplified to xtan⁡x+log⁡(cos⁡x)+C x \tan x + \log(\cos x) + C
step 3
The third integral is ∫x3+x2−5x+15(x2+5)(x2+2x+3)dx \int \frac{x^3 + x^2 - 5x + 15}{(x^2 + 5)(x^2 + 2x + 3)} dx . This can be expressed as 12(log⁡(x2+2x+3)−25tan⁡−1(x5)+52tan⁡−1(x+12))+C \frac{1}{2} \left( \log(x^2 + 2x + 3) - 2\sqrt{5} \tan^{-1}\left(\frac{x}{\sqrt{5}}\right) + 5\sqrt{2} \tan^{-1}\left(\frac{x + 1}{\sqrt{2}}\right) \right) + C
Answer
The first integral diverges, the second evaluates to xtan⁡x+log⁡(cos⁡x)+C x \tan x + \log(\cos x) + C , and the third evaluates to 12(log⁡(x2+2x+3)−25tan⁡−1(x5)+52tan⁡−1(x+12))+C \frac{1}{2} \left( \log(x^2 + 2x + 3) - 2\sqrt{5} \tan^{-1}\left(\frac{x}{\sqrt{5}}\right) + 5\sqrt{2} \tan^{-1}\left(\frac{x + 1}{\sqrt{2}}\right) \right) + C
Key Concept
Understanding convergence of integrals and integration techniques such as partial fractions and integration by parts
Explanation
The first integral diverges, while the others can be solved using standard integration techniques, yielding specific results.
solve e f g step by step
Let's solve the integrals step by step. (e) ∫1∞ln⁡xx dx \int_{1}^{\infty} \frac{\ln x}{x} \, dx 1. Evaluate the integral: The integral can be rewritten using the limit: ∫1∞ln⁡xx dx=lim⁡b→∞∫1bln⁡xx dx \int_{1}^{\infty} \frac{\ln x}{x} \, dx = \lim_{b \to \infty} \int_{1}^{b} \frac{\ln x}{x} \, dx 2. Use integration by parts: Let u=ln⁡x u = \ln x and dv=1x dx dv = \frac{1}{x} \, dx . Then, du=1x dx du = \frac{1}{x} \, dx and v=ln⁡x v = \ln x . ∫ln⁡x d(ln⁡x)=ln⁡2x/2 \int \ln x \, d(\ln x) = \ln^2 x / 2 3. Evaluate the limits: lim⁡b→∞[(ln⁡b)22−(ln⁡1)22]=lim⁡b→∞(ln⁡b)22 \lim_{b \to \infty} \left[ \frac{(\ln b)^2}{2} - \frac{(\ln 1)^2}{2} \right] = \lim_{b \to \infty} \frac{(\ln b)^2}{2} As b→∞ b \to \infty , (ln⁡b)22→∞ \frac{(\ln b)^2}{2} \to \infty . Therefore, the integral diverges. (f) ∫xcos⁡2x dx \int \frac{x}{\cos^2 x} \, dx 1. Recognize the integrand: The integrand can be rewritten as: ∫xsec⁡2x dx \int x \sec^2 x \, dx 2. Use integration by parts: Let u=x u = x and dv=sec⁡2x dx dv = \sec^2 x \, dx . Then, du=dx du = dx and v=tan⁡x v = \tan x . ∫xsec⁡2x dx=xtan⁡x−∫tan⁡x dx \int x \sec^2 x \, dx = x \tan x - \int \tan x \, dx 3. Integrate \( \tan x \): ∫tan⁡x dx=−ln⁡∣cos⁡x∣+C \int \tan x \, dx = -\ln |\cos x| + C 4. Combine results: ∫xcos⁡2x dx=xtan⁡x+ln⁡∣cos⁡x∣+C \int \frac{x}{\cos^2 x} \, dx = x \tan x + \ln |\cos x| + C (g) ∫x3+x2−5x+15(x2+5)(x2+2x+3) dx \int \frac{x^3 + x^2 - 5x + 15}{(x^2 + 5)(x^2 + 2x + 3)} \, dx 1. Use partial fraction decomposition: We express the integrand as: x3+x2−5x+15(x2+5)(x2+2x+3)=Ax+Bx2+5+Cx+Dx2+2x+3 \frac{x^3 + x^2 - 5x + 15}{(x^2 + 5)(x^2 + 2x + 3)} = \frac{Ax + B}{x^2 + 5} + \frac{Cx + D}{x^2 + 2x + 3} 2. Multiply through by the denominator and equate coefficients to find A,B,C,D A, B, C, D . 3. Integrate each term: After finding the coefficients, integrate each term separately. The integrals will typically involve logarithmic and arctangent functions. 4. Final result: The final result will be a combination of these integrals, which can be expressed as: ∫x3+x2−5x+15(x2+5)(x2+2x+3) dx=(result from partial fractions)+C \int \frac{x^3 + x^2 - 5x + 15}{(x^2 + 5)(x^2 + 2x + 3)} \, dx = \text{(result from partial fractions)} + C If you need further assistance with the partial fraction decomposition or any specific part of the calculations, feel free to ask!
give g more details
To solve the integral ∫x3+x2−5x+15(x2+5)(x2+2x+3) dx, \int \frac{x^3 + x^2 - 5x + 15}{(x^2 + 5)(x^2 + 2x + 3)} \, dx, we will use the method of partial fractions. Here are the steps in detail: Step 1: Factor the Denominator The denominator is already factored as (x2+5)(x2+2x+3)(x^2 + 5)(x^2 + 2x + 3). Step 2: Set Up Partial Fractions We express the integrand as a sum of partial fractions: x3+x2−5x+15(x2+5)(x2+2x+3)=Ax+Bx2+5+Cx+Dx2+2x+3. \frac{x^3 + x^2 - 5x + 15}{(x^2 + 5)(x^2 + 2x + 3)} = \frac{Ax + B}{x^2 + 5} + \frac{Cx + D}{x^2 + 2x + 3}. Step 3: Clear the Denominator Multiply both sides by the denominator (x2+5)(x2+2x+3)(x^2 + 5)(x^2 + 2x + 3): x3+x2−5x+15=(Ax+B)(x2+2x+3)+(Cx+D)(x2+5). x^3 + x^2 - 5x + 15 = (Ax + B)(x^2 + 2x + 3) + (Cx + D)(x^2 + 5). Step 4: Expand and Collect Like Terms Expand the right-hand side: Ax3+(2A+B)x2+(3A+2B)x+5Cx+(Cx2+5D). Ax^3 + (2A + B)x^2 + (3A + 2B)x + 5Cx + (Cx^2 + 5D). Combine like terms: (A+C)x3+(2A+B+C)x2+(3A+2B+5D)x+5D. (A + C)x^3 + (2A + B + C)x^2 + (3A + 2B + 5D)x + 5D. Step 5: Set Up a System of Equations Now, equate the coefficients from both sides: 1. A+C=1A + C = 1 (coefficient of x3x^3) 2. 2A+B+C=12A + B + C = 1 (coefficient of x2x^2) 3. 3A+2B+5C=−53A + 2B + 5C = -5 (coefficient of xx) 4. 5D=155D = 15 (constant term) From the fourth equation, we find D=3D = 3. Step 6: Substitute DD and Solve the System Substituting D=3D = 3 into the third equation gives: 3A+2B+5C=−5. 3A + 2B + 5C = -5. Now we have: 1. A+C=1A + C = 1 2. 2A+B+C=12A + B + C = 1 3. 3A+2B+5C=−53A + 2B + 5C = -5 From the first equation, we can express CC in terms of AA: C=1−A. C = 1 - A. Substituting CC into the second equation: 2A+B+(1−A)=1  ⟹  A+B=0  ⟹  B=−A. 2A + B + (1 - A) = 1 \implies A + B = 0 \implies B = -A. Now substitute BB and CC into the third equation: 3A+2(−A)+5(1−A)=−5  ⟹  3A−2A+5−5A=−5  ⟹  −4A+5=−5  ⟹  −4A=−10  ⟹  A=52. 3A + 2(-A) + 5(1 - A) = -5 \implies 3A - 2A + 5 - 5A = -5 \implies -4A + 5 = -5 \implies -4A = -10 \implies A = \frac{5}{2}. Now, substituting AA back to find BB and CC: B=−52,C=1−52=−32. B = -\frac{5}{2}, \quad C = 1 - \frac{5}{2} = -\frac{3}{2}. Step 7: Write the Partial Fraction Decomposition Now we have: x3+x2−5x+15(x2+5)(x2+2x+3)=52x−52x2+5+−32x+3x2+2x+3. \frac{x^3 + x^2 - 5x + 15}{(x^2 + 5)(x^2 + 2x + 3)} = \frac{\frac{5}{2}x - \frac{5}{2}}{x^2 + 5} + \frac{-\frac{3}{2}x + 3}{x^2 + 2x + 3}. Step 8: Integrate Each Term Now we can integrate each term separately: 1. For ∫52x−52x2+5 dx\int \frac{\frac{5}{2}x - \frac{5}{2}}{x^2 + 5} \, dx, we can split it into two integrals. 2. For ∫−32x+3x2+2x+3 dx\int \frac{-\frac{3}{2}x + 3}{x^2 + 2x + 3} \, dx, we can also split it. Final Integration After integrating each term, we will combine the results and add the constant of integration CC. If you need further assistance with the integration of each term, feel free
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