Question Number 222619 by ahmed2025 last updated on 01/Jul/25
Answered by wewji12 last updated on 02/Jul/25
$$\int\:\:\frac{\mathrm{2}}{\mathrm{2}−\mathrm{3}{x}}\:\:\mathrm{d}{x}=−\frac{\mathrm{1}}{\mathrm{3}}\mathrm{ln}\left(\mathrm{2}−\mathrm{3}{x}\right)+{C} \\ $$$$\int_{−\mathrm{1}} ^{\:\:\mathrm{2}} \:\frac{\mathrm{2}}{\mathrm{2}−\mathrm{3}{x}}\:\mathrm{d}{x}=\int_{\:−\mathrm{1}} ^{\frac{\mathrm{2}}{\mathrm{3}}} \frac{\mathrm{2}}{\mathrm{2}−\mathrm{3}{x}}\:\mathrm{d}{x}\:+\int_{\frac{\mathrm{2}}{\mathrm{3}}} ^{\:\mathrm{2}} \:\frac{\mathrm{2}}{\mathrm{2}−\mathrm{3}{x}}\:\mathrm{d}{x} \\ $$$$\int_{{a}} ^{{c}} +\int_{{c}} ^{{b}} =\infty−\infty \\ $$$${P}.{V}\int_{−\mathrm{1}} ^{\:\:\mathrm{2}} \:\:\frac{\mathrm{2}}{\mathrm{2}−\mathrm{3}{x}}\:\mathrm{d}{x}\:\left(\mathrm{Cauchy}−\mathrm{Principal}\:\mathrm{Value}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}\mathrm{ln}\left(\frac{\mathrm{5}}{\mathrm{4}}\right) \\ $$
Answered by profcedricjunior last updated on 01/Jul/25
![I=∫_(−1) ^2 (2/(2−3x))dx=−(2/3)∫_(−1) ^2 ((−3)/(2−3x))dx I=−(2/3)[ln(2−3x)]_(−1) ^2 =−(2/3)(ln(−4)−ln5) =((2ln5)/3)−((2ln4)/3)−((2𝛑i)/3)](https://www.tinkutara.com/question/Q222620.png)
$$\boldsymbol{{I}}=\int_{−\mathrm{1}} ^{\mathrm{2}} \frac{\mathrm{2}}{\mathrm{2}−\mathrm{3}\boldsymbol{{x}}}\boldsymbol{{dx}}=−\frac{\mathrm{2}}{\mathrm{3}}\int_{−\mathrm{1}} ^{\mathrm{2}} \frac{−\mathrm{3}}{\mathrm{2}−\mathrm{3}\boldsymbol{{x}}}\boldsymbol{{dx}} \\ $$$$\boldsymbol{{I}}=−\frac{\mathrm{2}}{\mathrm{3}}\left[\boldsymbol{{ln}}\left(\mathrm{2}−\mathrm{3}\boldsymbol{{x}}\right)\right]_{−\mathrm{1}} ^{\mathrm{2}} =−\frac{\mathrm{2}}{\mathrm{3}}\left(\boldsymbol{{ln}}\left(−\mathrm{4}\right)−\boldsymbol{{ln}}\mathrm{5}\right) \\ $$$$\:\:\:\:=\frac{\mathrm{2}\boldsymbol{{ln}}\mathrm{5}}{\mathrm{3}}−\frac{\mathrm{2}\boldsymbol{{ln}}\mathrm{4}}{\mathrm{3}}−\frac{\mathrm{2}\boldsymbol{\pi{i}}}{\mathrm{3}} \\ $$
Commented by mr W last updated on 05/Jul/25
![totally wrong! acc. to you we would get ∫_(−1) ^(−2) (dx/x)=[ln x]_(−1) ^(−2) =ln (−2)−ln (−1) but correctly ∫(dx/x)=ln ∣x∣+C, so correctly ∫_(−1) ^(−2) (dx/x)=[ln ∣x∣]_(−1) ^(−2) =ln (2)−ln (1)=ln 2 but in current case f(x)=(2/(2−3x)) is not continous in [−1, 2] at x=(2/3), you may not ignore this and simply take ∫_(−1) ^2 f(x)dx=[F(x)]_(−1) ^2 =F(2)−F(−1)](https://www.tinkutara.com/question/Q222692.png)
$${totally}\:{wrong}! \\ $$$${acc}.\:{to}\:{you}\:{we}\:{would}\:{get} \\ $$$$\int_{−\mathrm{1}} ^{−\mathrm{2}} \frac{{dx}}{{x}}=\left[\mathrm{ln}\:{x}\right]_{−\mathrm{1}} ^{−\mathrm{2}} =\mathrm{ln}\:\left(−\mathrm{2}\right)−\mathrm{ln}\:\left(−\mathrm{1}\right) \\ $$$${but}\:{correctly}\:\int\frac{{dx}}{{x}}=\mathrm{ln}\:\mid{x}\mid+{C},\:{so} \\ $$$${correctly} \\ $$$$\int_{−\mathrm{1}} ^{−\mathrm{2}} \frac{{dx}}{{x}}=\left[\mathrm{ln}\:\mid{x}\mid\right]_{−\mathrm{1}} ^{−\mathrm{2}} =\mathrm{ln}\:\left(\mathrm{2}\right)−\mathrm{ln}\:\left(\mathrm{1}\right)=\mathrm{ln}\:\mathrm{2} \\ $$$$ \\ $$$${but}\:{in}\:{current}\:{case} \\ $$$${f}\left({x}\right)=\frac{\mathrm{2}}{\mathrm{2}−\mathrm{3}{x}}\:{is}\:{not}\:{continous}\:{in} \\ $$$$\left[−\mathrm{1},\:\mathrm{2}\right]\:{at}\:{x}=\frac{\mathrm{2}}{\mathrm{3}},\:{you}\:{may}\:{not}\:{ignore}\: \\ $$$${this}\:{and}\:{simply}\:{take} \\ $$$$\:\int_{−\mathrm{1}} ^{\mathrm{2}} {f}\left({x}\right){dx}=\left[{F}\left({x}\right)\right]_{−\mathrm{1}} ^{\mathrm{2}} ={F}\left(\mathrm{2}\right)−{F}\left(−\mathrm{1}\right) \\ $$