Question Number 179853 by AKSHAYTHAKUR last updated on 03/Nov/22
$$\int\frac{\boldsymbol{{xdx}}}{\left(\boldsymbol{{x}}−\mathrm{1}\right)\left(\boldsymbol{{x}}−\mathrm{2}\right)} \\ $$
Commented by CElcedricjunior last updated on 04/Nov/22
![∫ ((xdx)/((x−1)(x−2)))=∫[(1/(x−2))+(1/((x−1)(x−2)))]dx =ln∣x−2∣+∫[−(1/(x−1))+(1/(x−2))]dx =ln∣x−2∣−ln∣x−1∣+ln∣x−1∣ =ln∣(((x−2)^2 )/(x−1))∣+cste =>∫((xdx)/((x−1)(x−2)))=ln∣(((x−2)^2 )/(x−1))∣+cste ..............le celebre cedric junior........](https://www.tinkutara.com/question/Q179885.png)
$$\int\:\frac{\boldsymbol{{xdx}}}{\left(\boldsymbol{{x}}−\mathrm{1}\right)\left(\boldsymbol{{x}}−\mathrm{2}\right)}=\int\left[\frac{\mathrm{1}}{\boldsymbol{{x}}−\mathrm{2}}+\frac{\mathrm{1}}{\left(\boldsymbol{{x}}−\mathrm{1}\right)\left(\boldsymbol{{x}}−\mathrm{2}\right)}\right]\boldsymbol{{dx}} \\ $$$$={ln}\mid\boldsymbol{{x}}−\mathrm{2}\mid+\int\left[−\frac{\mathrm{1}}{\boldsymbol{{x}}−\mathrm{1}}+\frac{\mathrm{1}}{\boldsymbol{{x}}−\mathrm{2}}\right]\boldsymbol{{dx}} \\ $$$$=\boldsymbol{{ln}}\mid\boldsymbol{{x}}−\mathrm{2}\mid−\boldsymbol{{ln}}\mid\boldsymbol{{x}}−\mathrm{1}\mid+\boldsymbol{{ln}}\mid\boldsymbol{{x}}−\mathrm{1}\mid \\ $$$$=\boldsymbol{{ln}}\mid\frac{\left(\boldsymbol{{x}}−\mathrm{2}\right)^{\mathrm{2}} }{\boldsymbol{{x}}−\mathrm{1}}\mid+\boldsymbol{{cste}} \\ $$$$=>\int\frac{\boldsymbol{{xdx}}}{\left(\boldsymbol{{x}}−\mathrm{1}\right)\left(\boldsymbol{{x}}−\mathrm{2}\right)}=\boldsymbol{{ln}}\mid\frac{\left(\boldsymbol{{x}}−\mathrm{2}\right)^{\mathrm{2}} }{\boldsymbol{{x}}−\mathrm{1}}\mid+\boldsymbol{{cste}}\: \\ $$$$…………..{le}\:{celebre}\:{cedric}\:{junior}…….. \\ $$
Commented by Frix last updated on 04/Nov/22
$$\mathrm{ln}\:{a}\:−\mathrm{ln}\:{b}\:=\mathrm{ln}\:\frac{{a}}{{b}}\:\neq\:\frac{\mathrm{ln}\:{a}}{\mathrm{ln}\:{b}} \\ $$$$\mathrm{your}\:\mathrm{answer}\:\mathrm{is}\:\mathrm{wrong} \\ $$
Answered by Frix last updated on 03/Nov/22
$$\int\frac{{x}}{\left({x}−\mathrm{1}\right)\left({x}−\mathrm{2}\right)}{dx}=\int\left(\frac{\mathrm{2}}{{x}−\mathrm{2}}−\frac{\mathrm{1}}{{x}−\mathrm{1}}\right){dx}= \\ $$$$=\mathrm{2ln}\:\mid{x}−\mathrm{2}\mid\:−\mathrm{ln}\:\mid{x}−\mathrm{1}\mid\:+{C} \\ $$
Commented by AKSHAYTHAKUR last updated on 09/Nov/22
$${thankyou}\:{ser} \\ $$