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Question Number 11619 by Nayon last updated on 29/Mar/17

Evaluate ∫(dx/((x+a)(x+b)))

$${Evaluate}\:\int\frac{{dx}}{\left({x}+{a}\right)\left({x}+{b}\right)} \\ $$

Commented by Nayon last updated on 29/Mar/17

someone pls ans

$${someone}\:{pls}\:{ans} \\ $$

Answered by mrW1 last updated on 29/Mar/17

=(1/(a−b))ln ∣((x+b)/(x+a))∣+C

$$=\frac{\mathrm{1}}{{a}−{b}}\mathrm{ln}\:\mid\frac{{x}+{b}}{{x}+{a}}\mid+{C} \\ $$

Commented by mrW1 last updated on 29/Mar/17

(1/((x+a)(x+b)))  =(p/(x+a))+(q/(x+b))=((px+pb+qx+qa)/((x+a)(x+b)))  ⇒p+q=0  ⇒pb+qa=1  −p=q=(1/(a−b))  ∫(1/((x+a)(x+b)))dx  =−(1/(a−b))∫(dx/(x+a))+(1/(a−b))∫(dx/(x+b))  =−(1/(a−b))ln ∣x+a∣+(1/(a−b))ln ∣x+b∣+C  =(1/(a−b))ln ∣((x+b)/(x+a))∣+C

$$\frac{\mathrm{1}}{\left({x}+{a}\right)\left({x}+{b}\right)} \\ $$$$=\frac{{p}}{{x}+{a}}+\frac{{q}}{{x}+{b}}=\frac{{px}+{pb}+{qx}+{qa}}{\left({x}+{a}\right)\left({x}+{b}\right)} \\ $$$$\Rightarrow{p}+{q}=\mathrm{0} \\ $$$$\Rightarrow{pb}+{qa}=\mathrm{1} \\ $$$$−{p}={q}=\frac{\mathrm{1}}{{a}−{b}} \\ $$$$\int\frac{\mathrm{1}}{\left({x}+{a}\right)\left({x}+{b}\right)}{dx} \\ $$$$=−\frac{\mathrm{1}}{{a}−{b}}\int\frac{{dx}}{{x}+{a}}+\frac{\mathrm{1}}{{a}−{b}}\int\frac{{dx}}{{x}+{b}} \\ $$$$=−\frac{\mathrm{1}}{{a}−{b}}\mathrm{ln}\:\mid{x}+{a}\mid+\frac{\mathrm{1}}{{a}−{b}}\mathrm{ln}\:\mid{x}+{b}\mid+{C} \\ $$$$=\frac{\mathrm{1}}{{a}−{b}}\mathrm{ln}\:\mid\frac{{x}+{b}}{{x}+{a}}\mid+{C} \\ $$

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