Question Number 158009 by zainaltanjung last updated on 30/Oct/21
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Answered by puissant last updated on 30/Oct/21
![Ω=∫_2 ^4 ((2x−1)/(x^2 −x))dx=[ln∣x^2 −x∣]_2 ^4 =ln12−ln2= ln(((12)/2))=ln6..](Q158019.png)
$$\Omega=\int_{\mathrm{2}} ^{\mathrm{4}} \frac{\mathrm{2}{x}−\mathrm{1}}{{x}^{\mathrm{2}} −{x}}{dx}=\left[{ln}\mid{x}^{\mathrm{2}} −{x}\mid\right]_{\mathrm{2}} ^{\mathrm{4}} \\ $$$$={ln}\mathrm{12}−{ln}\mathrm{2}=\:{ln}\left(\frac{\mathrm{12}}{\mathrm{2}}\right)={ln}\mathrm{6}.. \\ $$
Commented by zainaltanjung last updated on 30/Oct/21
$$\mathrm{ok}.\:\mathrm{right} \\ $$$$ \\ $$
Answered by Mr.D.N. last updated on 30/Oct/21
![∫_2 ^4 (( 2x−1)/(x^2 −x))dx Put, x^2 −x=t (2x−1)dx=dt t=12,2 ∫_2 ^( 12) (dt/t)= [ log t]_2 ^(12) or, log 12−log 2 or, log(2^2 .3)−log2 or, log2^2 +log3−log2 or, 2log2−log2 +log3 or, log 2+log 3= log(2.3) ∴ log 6 //.](Q158020.png)
$$\:\:\int_{\mathrm{2}} ^{\mathrm{4}} \:\frac{\:\mathrm{2x}−\mathrm{1}}{\mathrm{x}^{\mathrm{2}} −\mathrm{x}}\mathrm{dx} \\ $$$$\:\:\:\:\:\mathrm{Put},\:\mathrm{x}^{\mathrm{2}} −\mathrm{x}=\mathrm{t} \\ $$$$\:\:\:\:\:\left(\mathrm{2x}−\mathrm{1}\right)\mathrm{dx}=\mathrm{dt} \\ $$$$\:\:\:\:\mathrm{t}=\mathrm{12},\mathrm{2} \\ $$$$\:\:\:\int_{\mathrm{2}} ^{\:\mathrm{12}} \:\frac{\mathrm{dt}}{\mathrm{t}}=\:\left[\:\mathrm{log}\:\mathrm{t}\right]_{\mathrm{2}} ^{\mathrm{12}} \\ $$$$\:\mathrm{or},\:\:\:\mathrm{log}\:\mathrm{12}−\mathrm{log}\:\mathrm{2} \\ $$$$\:\:\mathrm{or},\:\:\mathrm{log}\left(\mathrm{2}^{\mathrm{2}} .\mathrm{3}\right)−\mathrm{log2} \\ $$$$\:\:\mathrm{or},\:\:\mathrm{log2}^{\mathrm{2}} +\mathrm{log3}−\mathrm{log2} \\ $$$$\:\mathrm{or},\:\:\mathrm{2log2}−\mathrm{log2}\:+\mathrm{log3} \\ $$$$\:\mathrm{or},\:\mathrm{log}\:\mathrm{2}+\mathrm{log}\:\mathrm{3}=\:\mathrm{log}\left(\mathrm{2}.\mathrm{3}\right) \\ $$$$\:\therefore\:\mathrm{log}\:\mathrm{6}\://. \\ $$
Commented by zainaltanjung last updated on 30/Oct/21
$$\mathrm{ok}.\:\mathrm{it}'\mathrm{right} \\ $$