Question Number 57228 by maxmathsup by imad last updated on 31/Mar/19
$${find}\:{f}\left({x}\right)\:=\int_{\mathrm{1}} ^{\mathrm{2}} \:\frac{{ln}\left(\mathrm{1}+{xt}\right)}{{t}^{\mathrm{2}} }\:{dt}\:\:{with}\:{x}>\mathrm{0} \\ $$
Commented by maxmathsup by imad last updated on 12/Apr/19
![by parts u^′ =(1/t^2 ) and v =ln(1+xt) ⇒f(x) =[−(1/t)ln(1+xt)]_1 ^2 + ∫_1 ^2 (1/t) (x/(1+xt))dt =ln(1+x)−(1/2)ln(1+2x) +∫_1 ^2 (x/(t(xt+1))) dt but ∫_1 ^2 (x/(t(xt +1))) dt =_(xt =u) ∫_x ^(2x) (x/((u/x)(u+1))) (du/x) =∫_x ^(2x) (x/(u(u+1))) dt =x ∫_x ^(2x) ((1/u) −(1/(u+1)))du =x [ln∣(u/(u+1))∣]_x ^(2x) =x {ln(((2x)/(2x+1)))−ln((x/(x+1)))} ⇒ f(x) =ln(1+x)−(1/2)ln(1+2x) +x ln(((2x)/(2x+1)))−xln((x/(x+1))) .](https://www.tinkutara.com/question/Q57787.png)
$${by}\:{parts}\:\:{u}^{'} \:=\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\:\:{and}\:{v}\:={ln}\left(\mathrm{1}+{xt}\right)\:\Rightarrow{f}\left({x}\right)\:=\left[−\frac{\mathrm{1}}{{t}}{ln}\left(\mathrm{1}+{xt}\right)\right]_{\mathrm{1}} ^{\mathrm{2}} \\ $$$$+\:\int_{\mathrm{1}} ^{\mathrm{2}} \:\:\frac{\mathrm{1}}{{t}}\:\frac{{x}}{\mathrm{1}+{xt}}{dt}\:={ln}\left(\mathrm{1}+{x}\right)−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{1}+\mathrm{2}{x}\right)\:\:+\int_{\mathrm{1}} ^{\mathrm{2}} \:\:\:\frac{{x}}{{t}\left({xt}+\mathrm{1}\right)}\:{dt}\:\:{but} \\ $$$$\int_{\mathrm{1}} ^{\mathrm{2}} \:\:\frac{{x}}{{t}\left({xt}\:+\mathrm{1}\right)}\:{dt}\:=_{{xt}\:={u}} \:\:\:\:\int_{{x}} ^{\mathrm{2}{x}} \:\:\frac{{x}}{\frac{{u}}{{x}}\left({u}+\mathrm{1}\right)}\:\frac{{du}}{{x}}\:=\int_{{x}} ^{\mathrm{2}{x}} \:\:\:\frac{{x}}{{u}\left({u}+\mathrm{1}\right)}\:{dt} \\ $$$$={x}\:\int_{{x}} ^{\mathrm{2}{x}} \left(\frac{\mathrm{1}}{{u}}\:−\frac{\mathrm{1}}{{u}+\mathrm{1}}\right){du}\:={x}\:\left[{ln}\mid\frac{{u}}{{u}+\mathrm{1}}\mid\right]_{{x}} ^{\mathrm{2}{x}} \:={x}\:\left\{{ln}\left(\frac{\mathrm{2}{x}}{\mathrm{2}{x}+\mathrm{1}}\right)−{ln}\left(\frac{{x}}{{x}+\mathrm{1}}\right)\right\}\:\Rightarrow \\ $$$${f}\left({x}\right)\:={ln}\left(\mathrm{1}+{x}\right)−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{1}+\mathrm{2}{x}\right)\:+{x}\:{ln}\left(\frac{\mathrm{2}{x}}{\mathrm{2}{x}+\mathrm{1}}\right)−{xln}\left(\frac{{x}}{{x}+\mathrm{1}}\right)\:. \\ $$