Question Number 47770 by tanmay.chaudhury50@gmail.com last updated on 14/Nov/18
$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{\mathrm{2}} −\mathrm{1}}{{lnx}}{dx} \\ $$
Commented by maxmathsup by imad last updated on 14/Nov/18
$${you}\:{are}\:{welcome}\:{sir}. \\ $$
Commented by maxmathsup by imad last updated on 14/Nov/18
![let I = ∫_0 ^1 ((x^2 −1)/(ln(x)))dx changement ln(x)=−t give x=e^(−t) I = ∫_0 ^(+∞) ((e^(−2t) −1)/(−t)) e^(−t) dt = ∫_0 ^∞ ((e^(−t) −e^(−3t) )/t) dt let determine f(x)=∫_0 ^∞ ((e^(−t) −e^(−3t) )/t) e^(−xt) dt with x≥0 we hsve f^′ (x)=−∫_0 ^∞ (e^(−t) −e^(−3t) )e^(−xt) dt =−∫_0 ^∞ ( e^(−(x+1)t) −e^(−(x+3)t) )dt =−[−(1/(x+1)) e^(−(x+1)t) +(1/(x+3)) e^(−(x+3)t) ]_(t=0) ^(+∞) =−((1/(x+1)) −(1/(x+3))) ⇒f(x)=−ln(((x+1)/(x+3)))+λ but ∃m>0 /∣f(x)∣≤m∫_0 ^∞ e^(−xt) dt =(m/x) →0 (x→+∞) ⇒ λ=lim_(x→+∞) (f(x)+ln(((x+1)/(x+3))))=0 ⇒f(x)=−ln(((x+1)/(x+3))) ⇒f(x)=ln(((x+3)/(x+1))) I =f(0)=ln(3) ⇒ ★ I =ln(3)★ .](https://www.tinkutara.com/question/Q47795.png)
$${let}\:{I}\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{x}^{\mathrm{2}} −\mathrm{1}}{{ln}\left({x}\right)}{dx}\:{changement}\:{ln}\left({x}\right)=−{t}\:{give}\:{x}=\overset{−{t}} {{e}} \\ $$$${I}\:=\:\int_{\mathrm{0}} ^{+\infty} \:\frac{{e}^{−\mathrm{2}{t}} −\mathrm{1}}{−{t}}\:{e}^{−{t}} {dt}\:=\:\int_{\mathrm{0}} ^{\infty} \:\frac{{e}^{−{t}} −{e}^{−\mathrm{3}{t}} }{{t}}\:{dt}\:\:{let}\:{determine} \\ $$$${f}\left({x}\right)=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{e}^{−{t}} −{e}^{−\mathrm{3}{t}} }{{t}}\:{e}^{−{xt}} {dt}\:{with}\:{x}\geqslant\mathrm{0}\:{we}\:{hsve}\: \\ $$$${f}^{'} \left({x}\right)=−\int_{\mathrm{0}} ^{\infty} \:\:\left({e}^{−{t}} −{e}^{−\mathrm{3}{t}} \right){e}^{−{xt}} {dt}\:=−\int_{\mathrm{0}} ^{\infty} \left(\:{e}^{−\left({x}+\mathrm{1}\right){t}} −{e}^{−\left({x}+\mathrm{3}\right){t}} \right){dt}\: \\ $$$$=−\left[−\frac{\mathrm{1}}{{x}+\mathrm{1}}\:{e}^{−\left({x}+\mathrm{1}\right){t}} \:+\frac{\mathrm{1}}{{x}+\mathrm{3}}\:{e}^{−\left({x}+\mathrm{3}\right){t}} \right]_{{t}=\mathrm{0}} ^{+\infty} =−\left(\frac{\mathrm{1}}{{x}+\mathrm{1}}\:−\frac{\mathrm{1}}{{x}+\mathrm{3}}\right)\:\Rightarrow{f}\left({x}\right)=−{ln}\left(\frac{{x}+\mathrm{1}}{{x}+\mathrm{3}}\right)+\lambda \\ $$$${but}\:\exists{m}>\mathrm{0}\:/\mid{f}\left({x}\right)\mid\leqslant{m}\int_{\mathrm{0}} ^{\infty} {e}^{−{xt}} {dt}\:=\frac{{m}}{{x}}\:\rightarrow\mathrm{0}\:\left({x}\rightarrow+\infty\right)\:\Rightarrow \\ $$$$\lambda={lim}_{{x}\rightarrow+\infty} \left({f}\left({x}\right)+{ln}\left(\frac{{x}+\mathrm{1}}{{x}+\mathrm{3}}\right)\right)=\mathrm{0}\:\Rightarrow{f}\left({x}\right)=−{ln}\left(\frac{{x}+\mathrm{1}}{{x}+\mathrm{3}}\right)\:\Rightarrow{f}\left({x}\right)={ln}\left(\frac{{x}+\mathrm{3}}{{x}+\mathrm{1}}\right) \\ $$$${I}\:={f}\left(\mathrm{0}\right)={ln}\left(\mathrm{3}\right)\:\Rightarrow\:\bigstar\:{I}\:={ln}\left(\mathrm{3}\right)\bigstar\:. \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 14/Nov/18
$${thank}\:{you}\:{sir}… \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 14/Nov/18
$${some}\:{tricks}\:{to}\:{solve}… \\ $$$${I}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{{a}} −\mathrm{1}}{{lnx}}{dx} \\ $$$$\frac{{dI}}{{da}}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\partial\left({x}^{{a}} −\mathrm{1}\right)}{\partial{a}}×\frac{\mathrm{1}}{{lnx}}{dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{{a}} ×{lnx}}{{lnx}}{dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{a}} {dx} \\ $$$$=\mid\frac{{x}^{{a}+\mathrm{1}} }{{a}+\mathrm{1}}\mid_{\mathrm{0}} ^{\mathrm{1}} =\frac{\mathrm{1}}{{a}+\mathrm{1}} \\ $$$$\frac{{dI}}{{da}}=\frac{\mathrm{1}}{{a}+\mathrm{1}} \\ $$$${dI}=\frac{{da}}{{a}+\mathrm{1}} \\ $$$${I}={ln}\left({a}+\mathrm{1}\right)+{C} \\ $$$${put}\:\:{a}=\mathrm{0}\:\:\:{so}\:\int\frac{{x}^{{a}} −\mathrm{1}}{{lnx}}{dx}\:\:=\mathrm{0} \\ $$$${hence}\:{I}={ln}\left({a}+\mathrm{1}\right)+{C} \\ $$$$\mathrm{0}={ln}\left(\mathrm{0}+\mathrm{1}\right)+{C} \\ $$$${C}=\mathrm{0} \\ $$$${so}\:{I}={ln}\left({a}+\mathrm{1}\right) \\ $$$${so}\:{answer}\:{for}\:\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{\mathrm{2}} −\mathrm{1}}{{lnx}}{dx}={ln}\left(\mathrm{2}+\mathrm{1}\right)={ln}\mathrm{3} \\ $$$$ \\ $$