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Answer 1

Answer:

$\mathrm{\bigg(\dfrac{ (c-b). ln(x-a) + (a-c).ln(x-b) + (b-a). ln(x-c)}{(a-b)(b-c)(c-a)} \bigg) + K}$

Step-by-step explanation:

Given :

$\int \dfrac{1}{(x-a)(x-b)(x-c)} \, dx$

To find :

Evaluate the integral

Solution :

First let us resolve this into partial fractions

• Resolving into partial fractions, we get,

$\dfrac{1}{(x-a)(x-b)(x-c)} = \\\\\\ \bigg( \dfrac{1}{(x-a)(a-b)(a-c)} + \dfrac{1}{(x-b)(b-a)(b-c)} + \dfrac{1}{(x-c)(c-a)(c-b)} \bigg)$

• Taking (a-b)(b-c)(c-a) common we get,

$\implies \bigg(\dfrac{1}{(a-b)(b-c)(c-a)} \bigg) \bigg( \dfrac{c-b}{x-a} + \dfrac{a-c}{x-b} + \dfrac{b-a}{x-c} \bigg)$

This is the fractional part of the given interval

• So, substituting in the integral we get,

$\implies \bigg(\dfrac{1}{(a-b)(b-c)(c-a)} \bigg){ \huge {\int} } \bigg( \dfrac{c-b}{x-a} + \dfrac{a-c}{x-b} + \dfrac{b-a}{x-c} \bigg) \, dx$

• Writing them separately

$\implies \bigg(\dfrac{1}{(a-b)(b-c)(c-a)} \bigg)\Bigg[ {\huge {\int}} \bigg( \dfrac{c-b}{x-a} \bigg) \, dx + { \huge {\int} } \bigg(\dfrac{a-c}{x-b} \bigg)\, dx + {\huge {\int}} \bigg( \dfrac{b-a}{x-c} \bigg) \, dx \Bigg]$

• Solving each separately,

We know that,

$\boxed {\mathrm { \large {\int} \dfrac{1}{x-a} \, dx = ln (x-a) + k \ ; \ k = constant}}$

So,

$\longmapsto {\huge {\int}} \bigg(\dfrac{c-b}{x-a} \bigg) \, dx = (c-b) . ln(x-a) + k$

$\longmapsto { \huge {\int}} \bigg(\dfrac{a-c}{x-b} \bigg) \, dx = (a-c) . ln(x-b) + k$

$\longmapsto { \huge {\int} }\bigg(\dfrac{b-a}{x-c} \bigg) \, dx = (b-a) . ln(x-c) + k$

Here, k = some constant

• Now rewriting the equations, we get,

$\implies \mathrm {\bigg(\dfrac{1}{(a-b)(b-c)(c-a)} \bigg) \Big( (c-b). ln(x-a) + (a-c).ln(x-b) + (b-a). ln(x-c) + K\Big)}$Thus, this is the required answer

$\implies \underline{\boxed {\mathbf {\bigg(\dfrac{ (c-b). ln(x-a) + (a-c).ln(x-b) + (b-a). ln(x-c)}{(a-b)(b-c)(c-a)} \bigg) } + K}}$

Here, K = constant

Hope it helps!!