Calculate the following integrals using partial fractions: 3.1 integral x3 + (5-5x)/(6x2+x-1) dx, 3.2 integral (2x3+6x2-12)/(x(x+3)(x2+3x+4)) dx

Mathematics
Calculate the following integrals using partial fractions: 3.1 integral x3 + (5-5x)/(6x2+x-1) dx, 3.2 integral (2x3+6x2-12)/(x(x+3)(x2+3x+4)) dx

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Answer

x44+23ln⁡∣3x−1∣−32ln⁡∣2x+1∣+C\frac{x^4}{4} + \frac{2}{3} \ln|3x-1| - \frac{3}{2} \ln|2x+1| + C

Step 1: Separate the integral into two parts. ∫x3+5−5x6x2+x−1dx=∫x3dx+∫5−5x6x2+x−1dx\int x^3 + \frac{5-5x}{6x^2+x-1} dx = \int x^3 dx + \int \frac{5-5x}{6x^2+x-1} dx

Step 2: Integrate the first term. ∫x3dx=x3+13+1+C1=x44+C1\int x^3 dx = \frac{x^{3+1}}{3+1} + C_1 = \frac{x^4}{4} + C_1

Step 3: Factor the denominator of the second term. The denominator is 6x2+x−16x^2+x-1. We look for two numbers that multiply to 6×(−1)=−66 \times (-1) = -6 and add to 11. These numbers are 33 and −2-2. 6x2+x−1=6x2+3x−2x−1=3x(2x+1)−1(2x+1)=(3x−1)(2x+1)6x^2+x-1 = 6x^2+3x-2x-1 = 3x(2x+1) - 1(2x+1) = (3x-1)(2x+1)

Step 4: Set up the partial fraction decomposition for the rational function. 5−5x(3x−1)(2x+1)=A3x−1+B2x+1\frac{5-5x}{(3x-1)(2x+1)} = \frac{A}{3x-1} + \frac{B}{2x+1} Multiply both sides by (3x−1)(2x+1)(3x-1)(2x+1): 5−5x=A(2x+1)+B(3x−1)5-5x = A(2x+1) + B(3x-1)

Step 5: Solve for the constants A and B. To find A, set 3x−1=0  ⟹  x=133x-1=0 \implies x=\frac{1}{3}: 5−5(13)=A(2(13)+1)+B(0)5-5\left(\frac{1}{3}\right) = A\left(2\left(\frac{1}{3}\right)+1\right) + B(0) 5−53=A(23+1)5-\frac{5}{3} = A\left(\frac{2}{3}+1\right) 15−53=A(53)\frac{15-5}{3} = A\left(\frac{5}{3}\right) 103=53A  ⟹  A=2\frac{10}{3} = \frac{5}{3}A \implies A = 2 To find B, set 2x+1=0  ⟹  x=−122x+1=0 \implies x=-\frac{1}{2}: 5−5(−12)=A(0)+B(3(−12)−1)5-5\left(-\frac{1}{2}\right) = A(0) + B\left(3\left(-\frac{1}{2}\right)-1\right) 5+52=B(−32−1)5+\frac{5}{2} = B\left(-\frac{3}{2}-1\right) 10+52=B(−52)\frac{10+5}{2} = B\left(-\frac{5}{2}\right) 152=−52B  ⟹  B=−3\frac{15}{2} = -\frac{5}{2}B \implies B = -3 So, the partial fraction decomposition is: 5−5x6x2+x−1=23x−1−32x+1\frac{5-5x}{6x^2+x-1} = \frac{2}{3x-1} - \frac{3}{2x+1}

Step 6: Integrate the partial fractions. ∫(23x−1−32x+1)dx=∫23x−1dx−∫32x+1dx\int \left(\frac{2}{3x-1} - \frac{3}{2x+1}\right) dx = \int \frac{2}{3x-1} dx - \int \frac{3}{2x+1} dx For ∫23x−1dx\int \frac{2}{3x-1} dx: Let u=3x−1u=3x-1, du=3dxdu=3dx. ∫2u13du=23∫1udu=23ln⁡∣u∣=23ln⁡∣3x−1∣\int \frac{2}{u} \frac{1}{3} du = \frac{2}{3} \int \frac{1}{u} du = \frac{2}{3} \ln|u| = \frac{2}{3} \ln|3x-1| For ∫32x+1dx\int \frac{3}{2x+1} dx: Let v=2x+1v=2x+1, dv=2dxdv=2dx. ∫3v12dv=32∫1vdv=32ln⁡∣v∣=32ln⁡∣2x+1∣\int \frac{3}{v} \frac{1}{2} dv = \frac{3}{2} \int \frac{1}{v} dv = \frac{3}{2} \ln|v| = \frac{3}{2} \ln|2x+1| Combining these, we get: ∫5−5x6x2+x−1dx=23ln⁡∣3x−1∣−32ln⁡∣2x+1∣+C2\int \frac{5-5x}{6x^2+x-1} dx = \frac{2}{3} \ln|3x-1| - \frac{3}{2} \ln|2x+1| + C_2

Step 7: Combine the results from Step 2 and Step 6. ∫x3+5−5x6x2+x−1dx=x44+23ln⁡∣3x−1∣−32ln⁡∣2x+1∣+C\int x^3 + \frac{5-5x}{6x^2+x-1} dx = \frac{x^4}{4} + \frac{2}{3} \ln|3x-1| - \frac{3}{2} \ln|2x+1| + C

The final answer is x44+23ln⁡∣3x−1∣−32ln⁡∣2x+1∣+C\boxed{\frac{x^4}{4} + \frac{2}{3} \ln|3x-1| - \frac{3}{2} \ln|2x+1| + C}.

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Separate the integral into two parts. x^3 + (5-5x)/(6x^2+x-1) dx = x^3 dx + (5-5x)/(6x^2+x-1) dx Step 2: Integrate the first term.

Calculate the following integrals using partial fractions: 3.1 integral x3 + (5-5x)/(6x2+x-1) dx, 3.2 integral (2x3+6x2-12)/(x(x+3)(x2+3x+4)) dx
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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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Step 1: Separate the integral into two parts. x^3 + (5-5x)/(6x^2+x-1) dx = x^3 dx + (5-5x)/(6x^2+x-1) dx Step 2: Integrate the first term. x^3 dx = x^3+13+1 + C_1 = (x^4)/(4) + C_1 Step 3: Factor the denominator of the second term. The denominator is 6x^2+x-1. We look for two numbers that multiply to 6 × (-1) = -6 and add to 1. These numbers are 3 and -2. 6x^2+x-1 = 6x^2+3x-2x-1 = 3x(2x+1) - 1(2x+1) = (3x-1)(2x+1) Step 4: Set up the partial fraction decomposition for the rational function. (5-5x)/((3x-1)(2x+1)) = (A)/(3x-1) + (B)/(2x+1) Multiply both sides by (3x-1)(2x+1): 5-5x = A(2x+1) + B(3x-1) Step 5: Solve for the constants A and B. To find A, set 3x-1=0 x=(1)/(3): 5-5((1)/(3)) = A(2((1)/(3))+1) + B(0) 5-(5)/(3) = A((2)/(3)+1) (15-5)/(3) = A((5)/(3)) (10)/(3) = (5)/(3)A A = 2 To find B, set 2x+1=0 x=-(1)/(2): 5-5(-(1)/(2)) = A(0) + B(3(-(1)/(2))-1) 5+(5)/(2) = B(-(3)/(2)-1) (10+5)/(2) = B(-(5)/(2)) (15)/(2) = -(5)/(2)B B = -3 So, the partial fraction decomposition is: (5-5x)/(6x^2+x-1) = (2)/(3x-1) - (3)/(2x+1) Step 6: Integrate the partial fractions. ((2)/(3x-1) - (3)/(2x+1)) dx = (2)/(3x-1) dx - (3)/(2x+1) dx For (2)/(3x-1) dx: Let u=3x-1, du=3dx. (2)/(u) (1)/(3) du = (2)/(3) (1)/(u) du = (2)/(3) |u| = (2)/(3) |3x-1| For (3)/(2x+1) dx: Let v=2x+1, dv=2dx. (3)/(v) (1)/(2) dv = (3)/(2) (1)/(v) dv = (3)/(2) |v| = (3)/(2) |2x+1| Combining these, we get: (5-5x)/(6x^2+x-1) dx = (2)/(3) |3x-1| - (3)/(2) |2x+1| + C_2 Step 7: Combine the results from Step 2 and Step 6. x^3 + (5-5x)/(6x^2+x-1) dx = (x^4)/(4) + (2)/(3) |3x-1| - (3)/(2) |2x+1| + C The final answer is (x^4)/(4) + (2)/(3) |3x-1| - (3)/(2) |2x+1| + C.