Question Number 67020 by mathmax by abdo last updated on 21/Aug/19
$${find}\:{f}\left({x}\right)\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{arctan}\left(\mathrm{1}+{xt}\right){dt}\:\:{with}\:{x}\:{real} \\ $$
Commented by mathmax by abdo last updated on 22/Aug/19
![f(x)=∫_0 ^1 arctan(1+xt)dt by parts f(x) =t arctan(1+xt)]_0 ^1 −∫_0 ^1 t (x/(1+(1+xt)^2 ))dt =arctan(1+x)−x ∫_0 ^1 (t/(1+(1+xt)^2 ))dt but ∫_0 ^1 (t/(1+(1+xt)^2 ))dt =∫_0 ^1 ((tdt)/(1+1+2xt +x^2 t^2 )) =∫_0 ^1 ((tdt)/(x^2 t^2 +2xt +2)) let decompose F(t)=(t/(x^2 t^2 +2xt+2)) Δ^′ =x^2 −2x^2 =−x^2 <0 if x≠0 so F(t) =(t/(x^2 (t^2 +((2t)/x)+(2/x^2 )))) =(t/(x^2 (t^2 +((2t)/x) +(1/x^2 ) +(2/x^2 )−(1/x^2 )))) =(t/(x^2 {(t+(1/x))^2 +(1/x^2 )})) we use the changement t+(1/x) =(1/(∣x∣))u ⇒xt+1=s(x)u ∫_(s(x)) ^((x+1)s(x)) (t/(1+(1+tx)^2 ))dt =∫_(s(x)) ^((x+1)s(x)) ((((1/(∣x∣))u−(1/x)))/(x^2 ×(1/x^2 )(1+u^2 )))du =(1/(∣x∣))∫_(s(x)) ^((x+1)s(x)) (u/(1+u^2 ))−(1/x) ∫_(s(x)) ^((x+1)s(x)) (du/(1+u^2 )) =(1/(2∣x∣))[ln(1+u^2 )]_(s(x)) ^((x+1)s(x)) −(1/x)[arctanu]_(s(x)) ^((x+1)s(x)) (s^2 (x)=1) =(1/(2∣x∣)){ln(1+(x+1)^2 )−ln2}−(1/x){arctan(x+1)s(x)−arctan(s(x))} ⇒f(x)= arctan(1+x)−((s(x))/2){ln(x^2 +2x+2)−ln(2)} +arctan(x+1)s(x)+x arctan(s(x)) with s(x) =1 if x>0 and s(x)=−1 if x<0](Q67080.png)
$${f}\left({x}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \:{arctan}\left(\mathrm{1}+{xt}\right){dt}\:\:{by}\:{parts}\: \\ $$$$\left.{f}\left({x}\right)\:={t}\:{arctan}\left(\mathrm{1}+{xt}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} \:−\int_{\mathrm{0}} ^{\mathrm{1}} \:{t}\:\frac{{x}}{\mathrm{1}+\left(\mathrm{1}+{xt}\right)^{\mathrm{2}} }{dt} \\ $$$$={arctan}\left(\mathrm{1}+{x}\right)−{x}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{t}}{\mathrm{1}+\left(\mathrm{1}+{xt}\right)^{\mathrm{2}} }{dt}\:{but} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{t}}{\mathrm{1}+\left(\mathrm{1}+{xt}\right)^{\mathrm{2}} }{dt}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{tdt}}{\mathrm{1}+\mathrm{1}+\mathrm{2}{xt}\:+{x}^{\mathrm{2}} {t}^{\mathrm{2}} }\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{tdt}}{{x}^{\mathrm{2}} {t}^{\mathrm{2}} \:+\mathrm{2}{xt}\:+\mathrm{2}} \\ $$$${let}\:{decompose}\:{F}\left({t}\right)=\frac{{t}}{{x}^{\mathrm{2}} {t}^{\mathrm{2}} \:+\mathrm{2}{xt}+\mathrm{2}} \\ $$$$\Delta^{'} \:={x}^{\mathrm{2}} −\mathrm{2}{x}^{\mathrm{2}} \:=−{x}^{\mathrm{2}} <\mathrm{0}\:\:{if}\:{x}\neq\mathrm{0}\:{so}\: \\ $$$${F}\left({t}\right)\:=\frac{{t}}{{x}^{\mathrm{2}} \left({t}^{\mathrm{2}} \:+\frac{\mathrm{2}{t}}{{x}}+\frac{\mathrm{2}}{{x}^{\mathrm{2}} }\right)}\:=\frac{{t}}{{x}^{\mathrm{2}} \left({t}^{\mathrm{2}} \:+\frac{\mathrm{2}{t}}{{x}}\:+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\:+\frac{\mathrm{2}}{{x}^{\mathrm{2}} }−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)} \\ $$$$=\frac{{t}}{{x}^{\mathrm{2}} \left\{\left({t}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right\}}\:\:\:{we}\:{use}\:{the}\:{changement}\:{t}+\frac{\mathrm{1}}{{x}}\:=\frac{\mathrm{1}}{\mid{x}\mid}{u}\:\Rightarrow{xt}+\mathrm{1}={s}\left({x}\right){u} \\ $$$$\int_{{s}\left({x}\right)} ^{\left({x}+\mathrm{1}\right){s}\left({x}\right)} \:\:\frac{{t}}{\mathrm{1}+\left(\mathrm{1}+{tx}\right)^{\mathrm{2}} }{dt}\:=\int_{{s}\left({x}\right)} ^{\left({x}+\mathrm{1}\right){s}\left({x}\right)} \:\:\frac{\left(\frac{\mathrm{1}}{\mid{x}\mid}{u}−\frac{\mathrm{1}}{{x}}\right)}{{x}^{\mathrm{2}} ×\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\left(\mathrm{1}+{u}^{\mathrm{2}} \right)}{du} \\ $$$$=\frac{\mathrm{1}}{\mid{x}\mid}\int_{{s}\left({x}\right)} ^{\left({x}+\mathrm{1}\right){s}\left({x}\right)} \:\:\frac{{u}}{\mathrm{1}+{u}^{\mathrm{2}} }−\frac{\mathrm{1}}{{x}}\:\int_{{s}\left({x}\right)} ^{\left({x}+\mathrm{1}\right){s}\left({x}\right)} \:\frac{{du}}{\mathrm{1}+{u}^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\mid{x}\mid}\left[{ln}\left(\mathrm{1}+{u}^{\mathrm{2}} \right)\right]_{{s}\left({x}\right)} ^{\left({x}+\mathrm{1}\right){s}\left({x}\right)} \:−\frac{\mathrm{1}}{{x}}\left[{arctanu}\right]_{{s}\left({x}\right)} ^{\left({x}+\mathrm{1}\right){s}\left({x}\right)} \:\:\:\:\:\left({s}^{\mathrm{2}} \left({x}\right)=\mathrm{1}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\mid{x}\mid}\left\{{ln}\left(\mathrm{1}+\left({x}+\mathrm{1}\right)^{\mathrm{2}} \right)−{ln}\mathrm{2}\right\}−\frac{\mathrm{1}}{{x}}\left\{{arctan}\left({x}+\mathrm{1}\right){s}\left({x}\right)−{arctan}\left({s}\left({x}\right)\right)\right\} \\ $$$$\Rightarrow{f}\left({x}\right)=\:{arctan}\left(\mathrm{1}+{x}\right)−\frac{{s}\left({x}\right)}{\mathrm{2}}\left\{{ln}\left({x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{2}\right)−{ln}\left(\mathrm{2}\right)\right\} \\ $$$$+{arctan}\left({x}+\mathrm{1}\right){s}\left({x}\right)+{x}\:{arctan}\left({s}\left({x}\right)\right) \\ $$$${with}\:{s}\left({x}\right)\:=\mathrm{1}\:{if}\:{x}>\mathrm{0}\:{and}\:{s}\left({x}\right)=−\mathrm{1}\:{if}\:{x}<\mathrm{0} \\ $$