91Ó°ÊÓ

We are given the integral to evaluate: $$\int \frac{x-5}{x^{2}(x+1)} d x$$

Now, we have to perform the partial fraction decomposition. We can rewrite the given fraction as: $$\frac{x-5}{x^{2}(x+1)} = \frac{A}{x} + \frac{B}{x^{2}} + \frac{C}{x+1}$$

To solve for A, B, and C, multiply both sides by the denominator \(x^2(x+1)\): $$(x-5) = A(x^2)(x+1) + B(x)(x+1) + C(x^2)$$ We can find A, B, and C by solving a system of equations. Let's substitute 3 different x-values, for example, x = -1, x = 0, and x = 1. For x = -1: $$(-6) = A(-1)(0) + B(-1)(0) + C(1) \Rightarrow C = -6$$ For x = 0: $$(-5)= A(0)(1) + B(0)(1) + C(0) \Rightarrow 0 = 0$$ For x = 1: $$(1-5) = A(1)(2) +B(1)(2) + C(1) \Rightarrow -4 = 2A + 2B -6$$ $$\Rightarrow A+B = 1$$ Since we cannot find A and B from the first two equations, we can take another value for x. Let's take x = 2: For x = 2: $$(2-5) = A(4)(3) + B(2)(3) + C(4) \Rightarrow -3 = 12A + 6B - 24$$ $$\Rightarrow 2A + B = 7$$ Now we got a system of two linear equations in A and B: $$A+B = 1$$ $$2A+B = 7$$ Subtracting the first equation from the second equation: $$A = 6$$ So we have found A, and we can find B by substituting A back into either equation: $$B = 1 - A = -5$$

Using the values of A, B, and C found above, we can rewrite the given integral as: $$\int \frac{x-5}{x^{2}(x+1)} d x = \int \left(\frac{6}{x} - \frac{5}{x^2} - \frac{6}{x+1}\right) dx$$

Now, we can integrate each of the terms separately: $$\int \frac{6}{x} dx = 6 \int \frac{1}{x} dx = 6 \ln |x| + C_1$$ $$\int \frac{-5}{x^2} dx = -5 \int x^{-2} dx = -5(-x^{-1}) + C_2 = 5x^{-1} + C_2$$ $$\int \frac{-6}{x+1} dx = -6 \int \frac{1}{x+1} dx = -6 \ln |x+1| + C_3$$

Now, we combine the individual integrals to obtain the final result. We can also combine the constants C_1, C_2, and C_3 into a single constant C: $$\int \frac{x-5}{x^{2}(x+1)} d x = 6 \ln |x| + 5x^{-1} - 6 \ln |x+1| + C$$