Question Number 98305 by mathmax by abdo last updated on 12/Jun/20
$$\mathrm{calculate}\:\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{ln}\left(\mathrm{x}\right)}{\left(\mathrm{1}+\mathrm{x}\right)^{\mathrm{3}} }\mathrm{dx} \\ $$
Answered by maths mind last updated on 13/Jun/20
![=∫_0 ^1 ((ln(t))/((1+t)^3 ))dt−∫_0 ^1 ((tln(t))/((1+t)^3 )) =2∫((ln(t))/((1+t)^3 ))dt−∫_0 ^1 ((ln(t))/((1+t)^2 ))=−(−2∫_0 ^1 ((ln(t)dt)/((1+t)^3 ))+∫_0 ^1 ((ln(t)dt)/((1+t)^2 )))=S ∫_y ^1 ((ln(t))/((a+t)^2 ))dt=[−((ln(t))/(a+t))]_y ^1 +∫_y ^1 (dt/(t(a+t))) =((ln(y))/(a+y))−(1/a){ln(y)+ln(1+a)−ln(a)} =((aln(y)−aln(y)−yln(y))/(a(a+y)))−((ln(1+a))/a)+((ln(a))/a) lim_(y→0) ∫_y ^1 ((ln(t))/((t+a)^2 ))dt=(1/a)ln((a/(a+1)))=f(a) f′(a)=−((ln((a/(a+1))))/a^2 )+(1/a)((1/a)−(1/(1+a))) S=−(f(1)+f′(1)) =−(ln((1/2))−ln((1/2))+(1/2))=−(1/2)](https://www.tinkutara.com/question/Q98321.png)
$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left({t}\right)}{\left(\mathrm{1}+{t}\right)^{\mathrm{3}} }{dt}−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{tln}\left({t}\right)}{\left(\mathrm{1}+{t}\right)^{\mathrm{3}} } \\ $$$$=\mathrm{2}\int\frac{{ln}\left({t}\right)}{\left(\mathrm{1}+{t}\right)^{\mathrm{3}} }{dt}−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left({t}\right)}{\left(\mathrm{1}+{t}\right)^{\mathrm{2}} }=−\left(−\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left({t}\right){dt}}{\left(\mathrm{1}+{t}\right)^{\mathrm{3}} }+\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left({t}\right){dt}}{\left(\mathrm{1}+{t}\right)^{\mathrm{2}} }\right)={S} \\ $$$$\int_{{y}} ^{\mathrm{1}} \frac{{ln}\left({t}\right)}{\left({a}+{t}\right)^{\mathrm{2}} }{dt}=\left[−\frac{{ln}\left({t}\right)}{{a}+{t}}\right]_{{y}} ^{\mathrm{1}} +\int_{{y}} ^{\mathrm{1}} \frac{{dt}}{{t}\left({a}+{t}\right)} \\ $$$$=\frac{{ln}\left({y}\right)}{{a}+{y}}−\frac{\mathrm{1}}{{a}}\left\{{ln}\left({y}\right)+{ln}\left(\mathrm{1}+{a}\right)−{ln}\left({a}\right)\right\} \\ $$$$=\frac{{aln}\left({y}\right)−{aln}\left({y}\right)−{yln}\left({y}\right)}{{a}\left({a}+{y}\right)}−\frac{{ln}\left(\mathrm{1}+{a}\right)}{{a}}+\frac{{ln}\left({a}\right)}{{a}} \\ $$$$\underset{{y}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:\int_{{y}} ^{\mathrm{1}} \frac{{ln}\left({t}\right)}{\left({t}+{a}\right)^{\mathrm{2}} }{dt}=\frac{\mathrm{1}}{{a}}{ln}\left(\frac{{a}}{{a}+\mathrm{1}}\right)={f}\left({a}\right) \\ $$$${f}'\left({a}\right)=−\frac{{ln}\left(\frac{{a}}{{a}+\mathrm{1}}\right)}{{a}^{\mathrm{2}} }+\frac{\mathrm{1}}{{a}}\left(\frac{\mathrm{1}}{{a}}−\frac{\mathrm{1}}{\mathrm{1}+{a}}\right) \\ $$$${S}=−\left({f}\left(\mathrm{1}\right)+{f}'\left(\mathrm{1}\right)\right) \\ $$$$=−\left({ln}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)−{ln}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)+\frac{\mathrm{1}}{\mathrm{2}}\right)=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$ \\ $$$$ \\ $$
Commented by abdomathmax last updated on 13/Jun/20
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir}\:\mathrm{mind} \\ $$
Answered by abdomathmax last updated on 14/Jun/20
![A =∫_0 ^∞ ((ln(x))/((1+x)^3 ))dx ⇒A=∫_0 ^1 ((ln(x))/((1+x)^3 ))dx +∫_1 ^(+∞) ((lnx)/((1+x)^3 ))dx(→x=(1/t)) =∫_0 ^1 ((lnx)/((1+x)^3 ))dx −∫_0 ^1 ((−lnt)/((1+(1/t))^3 ))(−(dt/t^2 )) =∫_0 ^1 ((lnx)/((1+x)^3 ))dx−∫_0 ^1 ((tlnt)/((1+t)^3 ))dt =∫_0 ^1 (((1−x)lnx)/((1+x)^3 )) dx =−∫_0 ^1 (((x+1−2)lnx)/((1+x)^3 )))dx =−∫_0 ^1 ((lnx)/((1+x)^2 ))dx +2 ∫_0 ^1 ((lnx)/((1+x)^3 ))dx =ln(2)+2 ∫_0 ^1 ((lnx)/((1+x)^3 ))dx by parts u^′ =(1+x)^(−3) ∫_0 ^1 ((ln(x))/((1+x)^3 ))dx =[((1/2)−(1/(2(1+x)^2 )))ln(x)]_0 ^1 −∫_0 ^1 ((1/2)−(1/(2(1+x)^2 )))(dx/x) =−(1/2) ∫_0 ^1 (((x^2 +2x)/(x(1+x)^2 )))dx =−(1/2) ∫_0 ^1 ((x+2)/((1+x)^2 ))dx =−(1/2) ∫_0 ^1 ((x+1+1)/((1+x)^2 ))dx =−(1/2) ∫_0 ^1 (dx/(1+x)) −(1/2) ∫_0 ^1 (dx/((1+x)^2 )) =−(1/2)ln(2)+(1/2)[(1/(x+1))]_0 ^1 =−((ln(2))/2) +(1/2)(−(1/2)) =−((ln(2))/2)−(1/4) ⇒ A =ln(2)+2(−((ln2)/2)−(1/4)) A=−(1/2)](https://www.tinkutara.com/question/Q98523.png)
$$\mathrm{A}\:=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{ln}\left(\mathrm{x}\right)}{\left(\mathrm{1}+\mathrm{x}\right)^{\mathrm{3}} }\mathrm{dx}\:\Rightarrow\mathrm{A}=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{ln}\left(\mathrm{x}\right)}{\left(\mathrm{1}+\mathrm{x}\right)^{\mathrm{3}} }\mathrm{dx}\: \\ $$$$+\int_{\mathrm{1}} ^{+\infty} \:\frac{\mathrm{lnx}}{\left(\mathrm{1}+\mathrm{x}\right)^{\mathrm{3}} }\mathrm{dx}\left(\rightarrow\mathrm{x}=\frac{\mathrm{1}}{\mathrm{t}}\right) \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{lnx}}{\left(\mathrm{1}+\mathrm{x}\right)^{\mathrm{3}} }\mathrm{dx}\:−\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{−\mathrm{lnt}}{\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{t}}\right)^{\mathrm{3}} }\left(−\frac{\mathrm{dt}}{\mathrm{t}^{\mathrm{2}} }\right) \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{lnx}}{\left(\mathrm{1}+\mathrm{x}\right)^{\mathrm{3}} }\mathrm{dx}−\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{tlnt}}{\left(\mathrm{1}+\mathrm{t}\right)^{\mathrm{3}} }\mathrm{dt} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\left(\mathrm{1}−\mathrm{x}\right)\mathrm{lnx}}{\left(\mathrm{1}+\mathrm{x}\right)^{\mathrm{3}} }\:\mathrm{dx}\:=−\int_{\mathrm{0}} ^{\mathrm{1}} \left(\frac{\left.\mathrm{x}+\mathrm{1}−\mathrm{2}\right)\mathrm{lnx}}{\left(\mathrm{1}+\mathrm{x}\right)^{\mathrm{3}} }\right)\mathrm{dx} \\ $$$$=−\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{lnx}}{\left(\mathrm{1}+\mathrm{x}\right)^{\mathrm{2}} }\mathrm{dx}\:+\mathrm{2}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{lnx}}{\left(\mathrm{1}+\mathrm{x}\right)^{\mathrm{3}} }\mathrm{dx} \\ $$$$=\mathrm{ln}\left(\mathrm{2}\right)+\mathrm{2}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{lnx}}{\left(\mathrm{1}+\mathrm{x}\right)^{\mathrm{3}} }\mathrm{dx}\:\:\mathrm{by}\:\mathrm{parts}\:\mathrm{u}^{'} =\left(\mathrm{1}+\mathrm{x}\right)^{−\mathrm{3}} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{ln}\left(\mathrm{x}\right)}{\left(\mathrm{1}+\mathrm{x}\right)^{\mathrm{3}} }\mathrm{dx}\:=\left[\left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{1}+\mathrm{x}\right)^{\mathrm{2}} }\right)\mathrm{ln}\left(\mathrm{x}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$−\int_{\mathrm{0}} ^{\mathrm{1}} \:\left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{1}+\mathrm{x}\right)^{\mathrm{2}} }\right)\frac{\mathrm{dx}}{\mathrm{x}} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \left(\frac{\mathrm{x}^{\mathrm{2}} \:+\mathrm{2x}}{\mathrm{x}\left(\mathrm{1}+\mathrm{x}\right)^{\mathrm{2}} }\right)\mathrm{dx} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{x}+\mathrm{2}}{\left(\mathrm{1}+\mathrm{x}\right)^{\mathrm{2}} }\mathrm{dx} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{x}+\mathrm{1}+\mathrm{1}}{\left(\mathrm{1}+\mathrm{x}\right)^{\mathrm{2}} }\mathrm{dx} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{dx}}{\mathrm{1}+\mathrm{x}}\:−\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{dx}}{\left(\mathrm{1}+\mathrm{x}\right)^{\mathrm{2}} } \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\left(\mathrm{2}\right)+\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{\mathrm{1}}{\mathrm{x}+\mathrm{1}}\right]_{\mathrm{0}} ^{\mathrm{1}} \:=−\frac{\mathrm{ln}\left(\mathrm{2}\right)}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{2}}\left(−\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$=−\frac{\mathrm{ln}\left(\mathrm{2}\right)}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{4}}\:\Rightarrow \\ $$$$\mathrm{A}\:=\mathrm{ln}\left(\mathrm{2}\right)+\mathrm{2}\left(−\frac{\mathrm{ln2}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{4}}\right) \\ $$$$\mathrm{A}=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$