1. Express x3(x+2)5x−2 in partial fraction.
Step 1: Set up the partial fraction decomposition.
The denominator has a repeated linear factor x3 and a distinct linear factor (x+2).
x3(x+2)5x−2=xA+x2B+x3C+x+2D
Step 2: Multiply both sides by the common denominator x3(x+2).
5x−2=Ax2(x+2)+Bx(x+2)+C(x+2)+Dx3
Step 3: Expand the right side and group terms by powers of x.
5x−2=A(x3+2x2)+B(x2+2x)+C(x+2)+Dx35x−2=Ax3+2Ax2+Bx2+2Bx+Cx+2C+Dx35x−2=(A+D)x3+(2A+B)x2+(2B+C)x+2C
Step 4: Equate coefficients of corresponding powers of x.
For x3: A+D=0(1)
For x2: 2A+B=0(2)
For x1: 2B+C=5(3)
For x0 (constant): 2C=−2(4)
Step 5: Solve the system of equations.
From (4): C=−1.
Substitute C=−1 into (3): 2B+(−1)=5⟹2B=6⟹B=3.
Substitute B=3 into (2): 2A+3=0⟹2A=−3⟹A=−23.
Substitute A=−23 into (1): −23+D=0⟹D=23.
Step 6: Write the partial fraction decomposition.
x3(x+2)5x−2=x−23+x23+x3−1+x+223x3(x+2)5x−2=−2x3+x23−x31+2(x+2)3
The final answer is −2x3+x23−x31+2(x+2)3.
2. Integrate ∫x3(x+2)5x−2dx.
Step 1: Use the partial fraction decomposition from the previous problem.
∫x3(x+2)5x−2dx=∫(−2x3+x23−x31+2(x+2)3)dx
Step 2: Integrate each term.
Recall that ∫xndx=n+1xn+1+K for n=−1, and ∫x1dx=ln∣x∣+K.
∫−2x3dx=−23ln∣x∣∫x23dx=∫3x−2dx=3−1x−1=−x3∫−x31dx=∫−x−3dx=−−2x−2=2x21∫2(x+2)3dx=23ln∣x+2∣
Step 3: Combine the results and add the constant of integration K.
−23ln∣x∣−x3+2x21+23ln∣x+2∣+K
This can be rewritten using logarithm properties lna−lnb=lnba:
23(ln∣x+2∣−ln∣x∣)−x3+2x21+K23lnxx+2−x3+2x21+K
The final answer is 23lnxx+2−x3+2x21+K.
3. Solve x−1x+1≥2.
Step 1: Move all terms to one side to get a single fraction.
x−1x+1−2≥0
Step 2: Find a common denominator and combine the terms.
x−1x+1−x−12(x−1)≥0x−1x+1−(2x−2)≥0x−1x+1−2x+2≥0x−1−x+3≥0
Step 3: Identify the critical points where the numerator or denominator is zero.
Numerator: −x+3=0⟹x=3
Denominator: x−1=0⟹x=1
Note that x=1 because the denominator cannot be zero.
Step 4: Test intervals defined by the critical points (−∞,1), (1,3], and [3,∞).
• For x<1 (e.g., x=0): 0−1−0+3=−13=−3. This is not ≥0.
• For 1<x≤3 (e.g., x=2): 2−1−2+3=11=1. This is ≥0. So, this interval is part of the solution.
• For x>3 (e.g., x=4): 4−1−4+3=3−1. This is not ≥0.
Step 5: Write the solution.
The inequality holds for 1<x≤3.
The final answer is 1<x≤3.
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This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
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Fresh day Modestine, let's solve. Here are the solutions to your questions: 1. Express (5x-2)/(x^3(x+2)) in partial fraction. Step 1: Set up the partial fraction decomposition. The denominator has a repeated linear factor x^3 and a distinct linear factor (x+2). (5x-2)/(x^3(x+2)) = (A)/(x) + (B)/(x^2) + (C)/(x^3) + (D)/(x+2) Step 2: Multiply both sides by the common denominator x^3(x+2). 5x-2 = A x^2(x+2) + B x(x+2) + C(x+2) + D x^3 Step 3: Expand the right side and group terms by powers of x. 5x-2 = A(x^3+2x^2) + B(x^2+2x) + C(x+2) + D x^3 5x-2 = Ax^3+2Ax^2 + Bx^2+2Bx + Cx+2C + Dx^3 5x-2 = (A+D)x^3 + (2A+B)x^2 + (2B+C)x + 2C Step 4: Equate coefficients of corresponding powers of x. For x^3: A+D = 0 (1) For x^2: 2A+B = 0 (2) For x^1: 2B+C = 5 (3) For x^0 (constant): 2C = -2 (4) Step 5: Solve the system of equations. From (4): C = -1. Substitute C=-1 into (3): 2B + (-1) = 5 2B = 6 B = 3. Substitute B=3 into (2): 2A + 3 = 0 2A = -3 A = -(3)/(2). Substitute A=-(3)/(2) into (1): -(3)/(2) + D = 0 D = (3)/(2). Step 6: Write the partial fraction decomposition. (5x-2)/(x^3(x+2)) = (-3)/(2)x + (3)/(x^2) + (-1)/(x^3) + (3)/(2)x+2 (5x-2)/(x^3(x+2)) = -(3)/(2x) + (3)/(x^2) - (1)/(x^3) + (3)/(2(x+2)) The final answer is -(3)/(2x) + (3)/(x^2) - (1)/(x^3) + (3)/(2(x+2)). 2. Integrate (5x-2)/(x^3(x+2))\,dx. Step 1: Use the partial fraction decomposition from the previous problem. (5x-2)/(x^3(x+2))\,dx = (-(3)/(2x) + (3)/(x^2) - (1)/(x^3) + (3)/(2(x+2)))\,dx Step 2: Integrate each term. Recall that x^n\,dx = x^n+1n+1 + K for n -1, and (1)/(x)\,dx = |x| + K. -(3)/(2x)\,dx = -(3)/(2)|x| (3)/(x^2)\,dx = 3x^-2\,dx = 3x^-1-1 = -(3)/(x) -(1)/(x^3)\,dx = -x^-3\,dx = -x^-2-2 = (1)/(2x^2) (3)/(2(x+2))\,dx = (3)/(2)|x+2| Step 3: Combine the results and add the constant of integration K. -(3)/(2)|x| - (3)/(x) + (1)/(2x^2) + (3)/(2)|x+2| + K This can be rewritten using logarithm properties a - b = (a)/(b): (3)/(2)(|x+2| - |x|) - (3)/(x) + (1)/(2x^2) + K (3)/(2)|(x+2)/(x)| - (3)/(x) + (1)/(2x^2) + K The final answer is (3)/(2)|(x+2)/(x)| - (3)/(x) + (1)/(2x^2) + K. 3. Solve (x+1)/(x-1) 2. Step 1: Move all terms to one side to get a single fraction. (x+1)/(x-1) - 2 0 Step 2: Find a common denominator and combine the terms. (x+1)/(x-1) - (2(x-1))/(x-1) 0 (x+1 - (2x-2))/(x-1) 0 (x+1 - 2x + 2)/(x-1) 0 (-x+3)/(x-1) 0 Step 3: Identify the critical points where the numerator or denominator is zero. Numerator: -x+3 = 0 x = 3 Denominator: x-1 = 0 x = 1 Note that x 1 because the denominator cannot be zero. Step 4: Test intervals defined by the critical points (-, 1), (1, 3], and [3, ). • For x < 1 (e.g., x=0): (-0+3)/(0-1) = (3)/(-1) = -3. This is not 0. • For 1 < x 3 (e.g., x=2): (-2+3)/(2-1) = (1)/(1) = 1. This is 0. So, this interval is part of the solution. • For x > 3 (e.g., x=4): (-4+3)/(4-1) = (-1)/(3). This is not 0. Step 5: Write the solution. The inequality holds for 1 < x 3. The final answer is 1 < x 3. What's next?