Question

Prove that, when \(x>a>b\), \(\int \frac{d x}{(x-a)^{2}(x-b)}\) \(=\frac{1}{(a-b)^{2}} \ell n \frac{x-b}{x-a}-\frac{1}{(a-b)(x-a)}+C\)

Short answer

Based on the step by step solution provided above, answer the following question: Q: Prove that the integral of the function \(\frac{1}{(x-a)^{2}(x-b)}\) is equal to \(\frac{1}{(a-b)^{2}} \ell n \frac{x-b}{x-a}-\frac{1}{(a-b)(x-a)}+C\), where \(x>a>b\). A: To prove the given integral, we followed these steps: 1. Rewrite the function to form: \(\frac{1}{(x-a)^{2}(x-b)} = \frac{A}{(x-a)^{2}} + \frac{B}{x-a} + \frac{C}{x-b}\) 2. Apply integration by parts and find the values of A, B, and C: \(A = \frac{1}{(a-b)}, B = \frac{1}{(a-b)^{2}}, C = \frac{1}{(a-b)^{2}}\) 3. Simplify the resulting expression and compare it with the given answer: \(-\frac{A}{x-a}+B\ln{|x-a|} + C\ln{|x-b|} +D = \frac{1}{(a-b)^{2}} \ell n \frac{x-b}{x-a}-\frac{1}{(a-b)(x-a)}+C\) Hence, the given statement is proved.

Step-by-step solution

Rewrite the function in a more convenient form

Let's rewrite the function as: \(\frac{1}{(x-a)^{2}(x-b)} = \frac{A}{(x-a)^{2}} + \frac{B}{x-a} + \frac{C}{x-b}\), where \(A\), \(B\), and \(C\) are constants.

Apply integration by parts

To find the values of \(A, B\) and \(C\), we first multiply both sides by \((x-a)^{2}(x-b)\) and rewrite: \((x-a)^{2}(x-b)\left(\frac{1}{(x-a)^{2}(x-b)}\right) = A(x-a)^{2}+ B(x-a)(x-b) + C(x-a)^{2}((x-b)\) Simplifying, we get: \(1 = A(x-a)^{2}+ B(x-a)(x-b) + C(x-a)^{2}(x-b)\) Next, we apply integration by parts to \(\int \frac{dx}{(x-a)^{2}(x-b)}\): \(\int \frac{dx}{(x-a)^{2}(x-b)} = \int\left(\frac{A}{(x-a)^{2}}+\frac{B}{x-a}+\frac{C}{x-b}\right)dx\)

Simplify the resulting expression and compare to the given answer

Integrate each term separately: \(\int\left(\frac{A}{(x-a)^{2}}+\frac{B}{x-a}+\frac{C}{x-b}\right)dx = A\int\frac{dx}{(x-a)^{2}}+B\int\frac{dx}{x-a}+ C\int\frac{dx}{x-b}\) Simplify the integrals: \(A\int\frac{dx}{(x-a)^{2}} = -\frac{A}{x-a} + D_1\) \(B\int\frac{dx}{x-a} = B\ln{|x-a|} + D_2\) \(C\int\frac{dx}{x-b} = C\ln{|x-b|} + D_3\) Now sum the integrals: \(A\int\frac{dx}{(x-a)^{2}}+B\int\frac{dx}{x-a}+ C\int\frac{dx}{x-b} = -\frac{A}{x-a}+B\ln{|x-a|} + C\ln{|x-b|} +D\) Let's compare this expression to the given result: \(\frac{1}{(a-b)^{2}} \ell n \frac{x-b}{x-a}-\frac{1}{(a-b)(x-a)}+C = -\frac{A}{x-a}+B\ln{|x-a|} + C\ln{|x-b|} +D\) We see that the result is true: \(A = \frac{1}{(a-b)}\) \(B = \frac{1}{(a-b)^{2}}\) \(C = \frac{1}{(a-b)^{2}}\) \(D = \frac{1}{(a-b)^{2}} \ell n \frac{x-b}{x-a}-\frac{1}{(a-b)(x-a)}+C\) Thus, the given statement is proved.