int {frac{{{x^2} + x - 6}}{{(x - 2)(x - 1)}}d | vedclass

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Question

(D) Given integral: $I = \int {\frac{{{x^2} + x - 6}}{{(x - 2)(x - 1)}}dx}$ Factorize the numerator: ${x^2} + x - 6 = (x + 3)(x - 2)$ Substitute this

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$x + 4\log (x - 1) + c$

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(D) Given integral: $I = \int {\frac{{{x^2} + x - 6}}{{(x - 2)(x - 1)}}dx}$ Factorize the numerator: ${x^2} + x - 6 = (x + 3)(x - 2)$ Substitute this into the integral: $I = \int {\frac{{(x + 3)(x - 2)}}{{(x - 2)(x - 1)}}dx} = \int {\frac{{x + 3}}{{x - 1}}dx}$ Rewrite the numerator: $x + 3 = (x - 1) + 4$ So,$I = \int {\frac{{(x - 1) + 4}}{{x - 1}}dx} = \int {\left( {1 + \frac{4}{{x - 1}}} \right)dx}$ Integrating term by term: $I = \int {1dx} + 4\int {\frac{1}{{x - 1}}dx} = x + 4\log |x - 1| + c$

$\int {\frac{{{x^2} + x - 6}}{{(x - 2)(x - 1)}}dx} = $

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