integrate 5³+17x-7 all over (x²+4)(x-1)dx

let's integrate this rational function using partial fraction decomposition. The integral is (5^3+17x-7)/((x^2+4)(x-1)) \, dx. Step 1: Simplify the numerator. 5^3+17x-7 = 125+17x-7 = 17x+118 So the integral becomes: (17x+118)/((x^2+4)(x-1)) \, dx Step 2: Set up the partial fraction decomposition. Since (x^2+4) is an irreducible quadratic factor and (x-1) is a linear factor, the decomposition takes the form: (17x+118)/((x^2+4)(x-1)) = (Ax+B)/(x^2+4) + (C)/(x-1) Step 3: Solve for the constants A, B, and C. Multiply both sides by (x^2+4)(x-1) to clear the denominators: 17x+118 = (Ax+B)(x-1) + C(x^2+4) Expand the right side: 17x+118 = Ax^2 - Ax + Bx - B + Cx^2 + 4C Group terms by powers of x: 17x+118 = (A+C)x^2 + (-A+B)x + (-B+4C) Equate the coefficients of corresponding powers of x on both sides: For x^2: A+C = 0 (*1) For x: -A+B = 17 (*2) For constant: -B+4C = 118 (*3) From (1), A = -C. Substitute this into (2): -(-C)+B = 17 C+B = 17 B = 17-C. Substitute B = 17-C into (*3): -(17-C)+4C = 118 -17+C+4C = 118 5C = 118+17 5C = 135 C = (135)/(5) = 27 Now find A and B: A = -C = -27. B = 17-C = 17-27 = -10. So the partial fraction decomposition is: (-27x-10)/(x^2+4) + (27)/(x-1) Step 4: Integrate each term. ( (-27x-10)/(x^2+4) + (27)/(x-1) ) dx = (-27x)/(x^2+4) \, dx - (10)/(x^2+4) \, dx + (27)/(x-1) \, dx For the first integral, (-27x)/(x^2+4) \, dx: Let u = x^2+4, then du = 2x \, dx, so x \, dx = (1)/(2) du. (-27)/(u) ((1)/(2) du) = -(27)/(2) (1)/(u) du = -(27)/(2) |u| = -(27)/(2) (x^2+4) (Since x^2+4 is always positive, we can remove the absolute value.) For the second integral, (10)/(x^2+4) \, dx: This is of the form (1)/(x^2+a^2) \, dx = (1)/(a) ((x)/(a)). Here a^2=4, so a=2. 10 (1)/(x^2+2^2) \, dx = 10 · (1)/(2) ((x)/(2)) = 5 ((x)/(2)) For the third integral, (27)/(x-1) \, dx: 27 (1)/(x-1) \, dx = 27 |x-1| Step 5: Combine the results. -(27)/(2) (x^2+4) - 5 ((x)/(2)) + 27 |x-1| + C The final answer is -(27)/(2) (x^2+4) - 5 ((x)/(2)) + 27 |x-1| + C. What's next? Send 'em! 📸