Question Number 163497 by Zaynal last updated on 07/Jan/22
Commented by Clide17 last updated on 07/Jan/22
![The function is discontinuous at interval [0, 3].](Q163509.png)
$$\mathrm{The}\:\mathrm{function}\:\mathrm{is}\:{discontinuous}\:\mathrm{at}\:\mathrm{interval}\:\left[\mathrm{0},\:\mathrm{3}\right]. \\ $$
Answered by TheSupreme last updated on 07/Jan/22
![x^3 +2x^2 +x−2=(x−2)(x^2 +1) (x/((x−2)(x^2 +1)))=(A/(x−2))+((Bx+C)/(x^2 +1))=((Ax^2 +A+Bx^2 +Cx−2Bx−2C)/((x^2 +1)(x−2)))=(((A+B)x^2 +(C−2B)x+A−2C)/(...))=(x/(...)) { ((A+B=0)),((C−2B=1)),((A−2C=0)) :} → { ((A=(2/5))),((B=−(2/5))),((C=(1/5))) :} (1/5)[∫_0 ^3 (2/(x−2))dx−∫_0 ^3 ((2x)/((x^2 +1)))dx+∫_0 ^3 (1/((x^2 +1)))dx]= =(1/5)[2ln∣x−2∣−ln(x^2 +1)+arctan(x)]∣_0 ^3 =−(2/5)ln(2)−(1/5)ln(10)+(1/5)arctan(3)](Q163499.png)
$${x}^{\mathrm{3}} +\mathrm{2}{x}^{\mathrm{2}} +{x}−\mathrm{2}=\left({x}−\mathrm{2}\right)\left({x}^{\mathrm{2}} +\mathrm{1}\right) \\ $$$$\frac{{x}}{\left({x}−\mathrm{2}\right)\left({x}^{\mathrm{2}} +\mathrm{1}\right)}=\frac{{A}}{{x}−\mathrm{2}}+\frac{{Bx}+{C}}{{x}^{\mathrm{2}} +\mathrm{1}}=\frac{{Ax}^{\mathrm{2}} +{A}+{Bx}^{\mathrm{2}} +{Cx}−\mathrm{2}{Bx}−\mathrm{2}{C}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)\left({x}−\mathrm{2}\right)}=\frac{\left({A}+{B}\right){x}^{\mathrm{2}} +\left({C}−\mathrm{2}{B}\right){x}+{A}−\mathrm{2}{C}}{...}=\frac{{x}}{...} \\ $$$$\begin{cases}{{A}+{B}=\mathrm{0}}\\{{C}−\mathrm{2}{B}=\mathrm{1}}\\{{A}−\mathrm{2}{C}=\mathrm{0}}\end{cases}\:\rightarrow\:\begin{cases}{{A}=\frac{\mathrm{2}}{\mathrm{5}}}\\{{B}=−\frac{\mathrm{2}}{\mathrm{5}}}\\{{C}=\frac{\mathrm{1}}{\mathrm{5}}}\end{cases} \\ $$$$\frac{\mathrm{1}}{\mathrm{5}}\left[\int_{\mathrm{0}} ^{\mathrm{3}} \frac{\mathrm{2}}{{x}−\mathrm{2}}{dx}−\int_{\mathrm{0}} ^{\mathrm{3}} \frac{\mathrm{2}{x}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)}{dx}+\int_{\mathrm{0}} ^{\mathrm{3}} \frac{\mathrm{1}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)}{dx}\right]= \\ $$$$=\frac{\mathrm{1}}{\mathrm{5}}\left[\mathrm{2}{ln}\mid{x}−\mathrm{2}\mid−{ln}\left({x}^{\mathrm{2}} +\mathrm{1}\right)+{arctan}\left({x}\right)\right]\mid_{\mathrm{0}} ^{\mathrm{3}} \\ $$$$=−\frac{\mathrm{2}}{\mathrm{5}}{ln}\left(\mathrm{2}\right)−\frac{\mathrm{1}}{\mathrm{5}}{ln}\left(\mathrm{10}\right)+\frac{\mathrm{1}}{\mathrm{5}}{arctan}\left(\mathrm{3}\right) \\ $$