Question Number 84021 by TANMAY PANACEA last updated on 08/Mar/20
Commented by abdomathmax last updated on 09/Mar/20
![I =∫_0 ^2 ((ln(1+2x))/(1+x^2 )) let f(a) =∫_0 ^2 ((ln(a+2x))/(1+x^2 ))dx we have f^′ (a) =(1/a)∫_0 ^2 (dx/((a+2x)(1+x^2 ))) let decompose F(x)=(1/((2x+a)(x^2 +1))) ⇒F(x)=(α/(2x+a)) +((βx+c)/(x^2 +1)) α =(1/(((−(a/2))^(2 ) +1))) =(1/((a^2 /4)+1)) =(4/(4+a^2 )) lim_(x→+∞) xF(x)=0=(α/2) +β ⇒β=−(α/2) =−(2/(4+a^2 )) F(0)=(1/a) =(α/a) +c ⇒c=(1/a)−(α/a) =((1−α)/a) =((1−(4/(4+a^2 )))/a) =(a^2 /(a(4+a^2 ))) ⇒ F(x)=(4/((a^2 +4)(2x+a))) +((((−2)/(a^2 +4))x +(a/(a^2 +4)))/(x^(2 ) +1)) =(1/(a^2 +4)){ (4/(2x+a)) +((−2x+a)/(x^2 +1))} ⇒ f^′ (a)=(1/(a(a^2 +4))){∫_0 ^2 ((4/(2x+a))+((−2x+a)/(x^2 +1)))dx} =(1/(a(a^2 +4))) ∫_0 ^2 ((2dx)/(x+(a/2))) −(2/(a(a^2 +4)))(1/2) ∫_0 ^2 ((2x−2a)/(x^2 +1))dx =(2/(a(a^2 +4)))[ln∣x+(a/2)∣]_0 ^2 −(1/(a(a^2 +4))) ∫_0 ^2 ((2xdx)/(x^2 +1)) +(1/((a^2 +4)))∫_0 ^2 (2/(x^2 +1))dx =(2/(a(a^2 +4)))(ln(2+(a/2))−ln((a/2)))−(1/(a(a^2 +4)))[ln(x^2 +1)]_0 ^2 +(2/(a^2 +4))[arctan(x)]_0 ^2 =(2/(a(a^2 +4)))ln(((4+a)/a))−((ln5)/(a(a^(2 ) +4))) +((2arctan(2))/(a^2 +4)) ⇒f(a)=∫_1 ^a (2/(t(t^2 +4)))ln(((t+4)/t))dt−ln(5)∫_1 ^a (dt/(t(t^2 +4))) +2arctan(2) ∫_1 ^a (dt/(t^2 +4)) +C ....be contonued...](https://www.tinkutara.com/question/Q84048.png)
$${I}\:=\int_{\mathrm{0}} ^{\mathrm{2}} \:\frac{{ln}\left(\mathrm{1}+\mathrm{2}{x}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }\:\:{let}\:{f}\left({a}\right)\:=\int_{\mathrm{0}} ^{\mathrm{2}} \:\frac{{ln}\left({a}+\mathrm{2}{x}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx} \\ $$$${we}\:{have}\:{f}^{'} \left({a}\right)\:=\frac{\mathrm{1}}{{a}}\int_{\mathrm{0}} ^{\mathrm{2}} \:\frac{{dx}}{\left({a}+\mathrm{2}{x}\right)\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}\:{let}\:{decompose} \\ $$$${F}\left({x}\right)=\frac{\mathrm{1}}{\left(\mathrm{2}{x}+{a}\right)\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)}\:\Rightarrow{F}\left({x}\right)=\frac{\alpha}{\mathrm{2}{x}+{a}}\:+\frac{\beta{x}+{c}}{{x}^{\mathrm{2}} \:+\mathrm{1}} \\ $$$$\alpha\:=\frac{\mathrm{1}}{\left(\left(−\frac{{a}}{\mathrm{2}}\right)^{\mathrm{2}\:} +\mathrm{1}\right)}\:=\frac{\mathrm{1}}{\frac{{a}^{\mathrm{2}} }{\mathrm{4}}+\mathrm{1}}\:=\frac{\mathrm{4}}{\mathrm{4}+{a}^{\mathrm{2}} } \\ $$$${lim}_{{x}\rightarrow+\infty} \:\:\:{xF}\left({x}\right)=\mathrm{0}=\frac{\alpha}{\mathrm{2}}\:+\beta\:\Rightarrow\beta=−\frac{\alpha}{\mathrm{2}}\:=−\frac{\mathrm{2}}{\mathrm{4}+{a}^{\mathrm{2}} } \\ $$$${F}\left(\mathrm{0}\right)=\frac{\mathrm{1}}{{a}}\:=\frac{\alpha}{{a}}\:+{c}\:\Rightarrow{c}=\frac{\mathrm{1}}{{a}}−\frac{\alpha}{{a}}\:=\frac{\mathrm{1}−\alpha}{{a}} \\ $$$$=\frac{\mathrm{1}−\frac{\mathrm{4}}{\mathrm{4}+{a}^{\mathrm{2}} }}{{a}}\:=\frac{{a}^{\mathrm{2}} }{{a}\left(\mathrm{4}+{a}^{\mathrm{2}} \right)}\:\Rightarrow \\ $$$${F}\left({x}\right)=\frac{\mathrm{4}}{\left({a}^{\mathrm{2}} \:+\mathrm{4}\right)\left(\mathrm{2}{x}+{a}\right)}\:+\frac{\frac{−\mathrm{2}}{{a}^{\mathrm{2}} +\mathrm{4}}{x}\:+\frac{{a}}{{a}^{\mathrm{2}} \:+\mathrm{4}}}{{x}^{\mathrm{2}\:} +\mathrm{1}} \\ $$$$=\frac{\mathrm{1}}{{a}^{\mathrm{2}} +\mathrm{4}}\left\{\:\frac{\mathrm{4}}{\mathrm{2}{x}+{a}}\:+\frac{−\mathrm{2}{x}+{a}}{{x}^{\mathrm{2}} \:+\mathrm{1}}\right\}\:\Rightarrow \\ $$$${f}^{'} \left({a}\right)=\frac{\mathrm{1}}{{a}\left({a}^{\mathrm{2}} \:+\mathrm{4}\right)}\left\{\int_{\mathrm{0}} ^{\mathrm{2}} \:\left(\frac{\mathrm{4}}{\mathrm{2}{x}+{a}}+\frac{−\mathrm{2}{x}+{a}}{{x}^{\mathrm{2}} \:+\mathrm{1}}\right){dx}\right\} \\ $$$$=\frac{\mathrm{1}}{{a}\left({a}^{\mathrm{2}} \:+\mathrm{4}\right)}\:\int_{\mathrm{0}} ^{\mathrm{2}} \:\frac{\mathrm{2}{dx}}{{x}+\frac{{a}}{\mathrm{2}}}\:−\frac{\mathrm{2}}{{a}\left({a}^{\mathrm{2}} +\mathrm{4}\right)}\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\mathrm{2}} \:\:\frac{\mathrm{2}{x}−\mathrm{2}{a}}{{x}^{\mathrm{2}} \:+\mathrm{1}}{dx} \\ $$$$=\frac{\mathrm{2}}{{a}\left({a}^{\mathrm{2}} +\mathrm{4}\right)}\left[{ln}\mid{x}+\frac{{a}}{\mathrm{2}}\mid\right]_{\mathrm{0}} ^{\mathrm{2}} −\frac{\mathrm{1}}{{a}\left({a}^{\mathrm{2}} \:+\mathrm{4}\right)}\:\int_{\mathrm{0}} ^{\mathrm{2}} \:\frac{\mathrm{2}{xdx}}{{x}^{\mathrm{2}} \:+\mathrm{1}} \\ $$$$+\frac{\mathrm{1}}{\left({a}^{\mathrm{2}} \:+\mathrm{4}\right)}\int_{\mathrm{0}} ^{\mathrm{2}} \:\:\frac{\mathrm{2}}{{x}^{\mathrm{2}} +\mathrm{1}}{dx} \\ $$$$=\frac{\mathrm{2}}{{a}\left({a}^{\mathrm{2}} \:+\mathrm{4}\right)}\left({ln}\left(\mathrm{2}+\frac{{a}}{\mathrm{2}}\right)−{ln}\left(\frac{{a}}{\mathrm{2}}\right)\right)−\frac{\mathrm{1}}{{a}\left({a}^{\mathrm{2}} \:+\mathrm{4}\right)}\left[{ln}\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)\right]_{\mathrm{0}} ^{\mathrm{2}} \\ $$$$+\frac{\mathrm{2}}{{a}^{\mathrm{2}} \:+\mathrm{4}}\left[{arctan}\left({x}\right)\right]_{\mathrm{0}} ^{\mathrm{2}} \\ $$$$=\frac{\mathrm{2}}{{a}\left({a}^{\mathrm{2}} \:+\mathrm{4}\right)}{ln}\left(\frac{\mathrm{4}+{a}}{{a}}\right)−\frac{{ln}\mathrm{5}}{{a}\left({a}^{\mathrm{2}\:} +\mathrm{4}\right)}\:+\frac{\mathrm{2}{arctan}\left(\mathrm{2}\right)}{{a}^{\mathrm{2}} \:+\mathrm{4}} \\ $$$$\Rightarrow{f}\left({a}\right)=\int_{\mathrm{1}} ^{{a}} \:\frac{\mathrm{2}}{{t}\left({t}^{\mathrm{2}} \:+\mathrm{4}\right)}{ln}\left(\frac{{t}+\mathrm{4}}{{t}}\right){dt}−{ln}\left(\mathrm{5}\right)\int_{\mathrm{1}} ^{{a}} \:\frac{{dt}}{{t}\left({t}^{\mathrm{2}} \:+\mathrm{4}\right)} \\ $$$$+\mathrm{2}{arctan}\left(\mathrm{2}\right)\:\int_{\mathrm{1}} ^{{a}} \:\:\frac{{dt}}{{t}^{\mathrm{2}} \:+\mathrm{4}}\:+{C}\:….{be}\:{contonued}… \\ $$
Answered by TANMAY PANACEA last updated on 08/Mar/20
$$\left.\mathrm{1}\right)\int_{\mathrm{1}} ^{\infty} \frac{\mathrm{2}{x}^{\mathrm{3}} −\mathrm{1}}{{x}^{\mathrm{6}} +\mathrm{2}{x}^{\mathrm{3}} +\mathrm{9}{x}^{\mathrm{2}} +\mathrm{1}}{dx} \\ $$$$=\int_{\mathrm{1}} ^{\infty} \frac{\mathrm{2}{x}^{\mathrm{3}} −\mathrm{1}}{\left({x}^{\mathrm{3}} +\mathrm{1}\right)^{\mathrm{2}} +\mathrm{9}{x}^{\mathrm{2}} }{dx} \\ $$$$=\int_{\mathrm{1}} ^{\infty} \frac{\mathrm{2}{x}−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }}{\left(\frac{{x}^{\mathrm{3}} +\mathrm{1}}{{x}}\right)^{\mathrm{2}} +\mathrm{9}}{dx} \\ $$$$\int_{\mathrm{1}} ^{\infty} \frac{\mathrm{2}{x}−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }}{\left({x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} }{dx}=\int_{\mathrm{1}} ^{\infty} \frac{{d}\left({x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}}\right)}{\left({x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} } \\ $$$$\frac{\mathrm{1}}{\mathrm{3}}×\mid{tan}^{−\mathrm{1}} \left(\frac{\left.{x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}}\right)}{\mathrm{3}}\right)\mid_{\mathrm{1}} ^{\infty} \\ $$$$\frac{\mathrm{1}}{\mathrm{3}}×\left({tan}^{−\mathrm{1}} \infty−{tan}^{−\mathrm{1}} \frac{\mathrm{2}}{\mathrm{3}}\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{3}}×\left(\frac{\pi}{\mathrm{2}}−{tan}^{−\mathrm{1}} \frac{\mathrm{2}}{\mathrm{3}}\right) \\ $$
Answered by TANMAY PANACEA last updated on 08/Mar/20
Answered by TANMAY PANACEA last updated on 08/Mar/20
![∫_0 ^(π/4) ((ln(cotx))/([(sinx)^(2009) +(cosx)^(2009) ]^2 ))(sin2x)^(2008) ∫_0 ^(π/4) ((ln(cotx))/((sinx)^(4018) [1+(cotx)^(2009) ]^2 ))×2^(2008) ×sin^(2008) xcos^(2008) x ∫_0 ^(π/4) ((ln(cotx))/([1+(cotx)^(2009) ]^2 ))×2^(2008) ×((cos^(2008) x)/(sin^(2010) ))dx ∫_0 ^(π/4) ((ln(cotx))/([1+(cotx)^(2009) ]^2 ))×2^(2008) ×(cotx)^(2008) ×cosec^2 xdx t=cotx (dt/dx)=−cozec^2 x ∫_∞ ^1 ((lnt)/((1+t^(2009) )^2 ))×2^(2008) ×t^(2008) ×−dt 2^(2008) ∫_1 ^∞ ((lnt)/((1+t^(2009) )^2 ))×t^(2008) ×dt](https://www.tinkutara.com/question/Q84040.png)
$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{{ln}\left({cotx}\right)}{\left[\left({sinx}\right)^{\mathrm{2009}} +\left({cosx}\right)^{\mathrm{2009}} \right]^{\mathrm{2}} }\left({sin}\mathrm{2}{x}\right)^{\mathrm{2008}} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{{ln}\left({cotx}\right)}{\left({sinx}\right)^{\mathrm{4018}} \left[\mathrm{1}+\left({cotx}\right)^{\mathrm{2009}} \right]^{\mathrm{2}} }×\mathrm{2}^{\mathrm{2008}} ×{sin}^{\mathrm{2008}} {xcos}^{\mathrm{2008}} {x} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{{ln}\left({cotx}\right)}{\left[\mathrm{1}+\left({cotx}\right)^{\mathrm{2009}} \right]^{\mathrm{2}} }×\mathrm{2}^{\mathrm{2008}} ×\frac{{cos}^{\mathrm{2008}} {x}}{{sin}^{\mathrm{2010}} }{dx} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{{ln}\left({cotx}\right)}{\left[\mathrm{1}+\left({cotx}\right)^{\mathrm{2009}} \right]^{\mathrm{2}} }×\mathrm{2}^{\mathrm{2008}} ×\left({cotx}\right)^{\mathrm{2008}} ×{cosec}^{\mathrm{2}} {xdx} \\ $$$${t}={cotx}\:\:\:\frac{{dt}}{{dx}}=−{cozec}^{\mathrm{2}} {x} \\ $$$$\int_{\infty} ^{\mathrm{1}} \frac{{lnt}}{\left(\mathrm{1}+{t}^{\mathrm{2009}} \right)^{\mathrm{2}} }×\mathrm{2}^{\mathrm{2008}} ×{t}^{\mathrm{2008}} ×−{dt} \\ $$$$\mathrm{2}^{\mathrm{2008}} \int_{\mathrm{1}} ^{\infty} \frac{{lnt}}{\left(\mathrm{1}+{t}^{\mathrm{2009}} \right)^{\mathrm{2}} }×{t}^{\mathrm{2008}} ×{dt} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$