Question Number 11619 by Nayon last updated on 29/Mar/17
$${Evaluate}\:\int\frac{{dx}}{\left({x}+{a}\right)\left({x}+{b}\right)} \\ $$
Commented by Nayon last updated on 29/Mar/17
$${someone}\:{pls}\:{ans} \\ $$
Answered by mrW1 last updated on 29/Mar/17
$$=\frac{\mathrm{1}}{{a}−{b}}\mathrm{ln}\:\mid\frac{{x}+{b}}{{x}+{a}}\mid+{C} \\ $$
Commented by mrW1 last updated on 29/Mar/17
$$\frac{\mathrm{1}}{\left({x}+{a}\right)\left({x}+{b}\right)} \\ $$$$=\frac{{p}}{{x}+{a}}+\frac{{q}}{{x}+{b}}=\frac{{px}+{pb}+{qx}+{qa}}{\left({x}+{a}\right)\left({x}+{b}\right)} \\ $$$$\Rightarrow{p}+{q}=\mathrm{0} \\ $$$$\Rightarrow{pb}+{qa}=\mathrm{1} \\ $$$$−{p}={q}=\frac{\mathrm{1}}{{a}−{b}} \\ $$$$\int\frac{\mathrm{1}}{\left({x}+{a}\right)\left({x}+{b}\right)}{dx} \\ $$$$=−\frac{\mathrm{1}}{{a}−{b}}\int\frac{{dx}}{{x}+{a}}+\frac{\mathrm{1}}{{a}−{b}}\int\frac{{dx}}{{x}+{b}} \\ $$$$=−\frac{\mathrm{1}}{{a}−{b}}\mathrm{ln}\:\mid{x}+{a}\mid+\frac{\mathrm{1}}{{a}−{b}}\mathrm{ln}\:\mid{x}+{b}\mid+{C} \\ $$$$=\frac{\mathrm{1}}{{a}−{b}}\mathrm{ln}\:\mid\frac{{x}+{b}}{{x}+{a}}\mid+{C} \\ $$